Solution: (a) The highest order derivative involved is d2y which is the 2nd order derivative. Hence order of the differential equation is
dx 2
Example: 3
Solution: (d) 2. Making the above equation free from radical, as far as the derivatives are concerned, we have
Example: 4
Solution: (c) d 2 y + 1 3 = − dy i.e. d 2 y + x 1 3 + dy = 0
Example: 5 dx 2 dx dx 2 4 dx
Solution: (a) x4
Example: 6 The exponent of highest order derivative d2y will be 3. Hence degree of the differential equation is 3.
Solution: (c) dx 2
Example: 7 The degree of the differential equation d2y + 3 dy 2 = x2 log d2y is
Solution: (a) dx 2 dx dx 2
Example: 8
Solution: (b) (a) 1 (b) 2 (c) 3 (d) None of these
The above equation cannot be written as a polynomial in derivatives due to the term x2 log d2 y . Hence degree of the
dx
2
differential equation is ‘not defined’.
The order of the differential equation whose solution is x 2 + y 2 + 2gx + 2 fy + c = 0, is [MP PET 1995]
(a) 1 (b) 2 (c) 3 (d) 4
To eliminate the arbitrary constants g, f and c, we need 3 more equations, that by differentiating the equation 3 times.
Hence highest order derivative will be d3y . Hence order of the differential equation will be 3.
dx 3.
The order of the differential equation of all circles of radius r, having centre on y-axis and passing through the origin is
(a) 1 (b) 2 (c) 3 (d) 4
The equation of the circle will be x 2 + (y ± r)2 = r 2 . Hence the constant to be eliminated is r.
∴ order of differential equation =1
Y Y
(0, r) OX
(0, – r)
OX
The order of the differential equation, whose general solution is y = c1e x + c2e2x + c3e3x + c4e x+c5 , where c1,c2,c3,c4,c5
are arbitrary constants is
(a) 5 (b) 4 (c) 3 (d) None of these
Rewriting the given general solution, we have
y = c1e x + c 2e 2x + c3e 3x + c4 e x .e c5
= (c1 + c 4 .e c5 ) e x + c 2e 2x + c 3 e 3x = c1′ e x + c 2e 2x + c3e 3x
where c1′ = c1 + c 4 .e c5 . So there are 3 arbitrary constant associated with different terms. Hence the order of the
differential equation formed, will be 3.
The degree of the differential equation satisfying 1 − x2 + 1 − y 2 = a(x − y) is
(a) 1 (b) 2 (c) 3 (d) None of these
To eliminate a the above equation is differentiated once and exponent of dy will be1. Hence degree is 1
dx
The order and degree of y =1+ dy + 1 dy 2 + 1 dy 3 + ....... is
dx 2! dx 3! dx
(a) 1, 2 (b) 1, 1
(c) Order 1, degree not defined (d) None of these
The given differential equation can be re-written as dy ⇒ ln y = dy
dx
y = e dx
This is a polynomial in derivative. Hence order is 1 and degree 1.
Example: 9 The order and degree of d2y = sin dy + x is
Solution: (b) dx 2 dx
(a) 2, 1 (b) Order 2, degree not defined
(c) 2, 0 (d) None of these
As the highest order derivative involved is d2y . Hence order is 2.
dx 2
The given differential equation cannot be written as a polynomial in derivatives, the degree is not defined.
Formation of Differential Equation .
Formulating a differential equation from a given equation representing a family of curves means finding a
differential equation whose solution is the given equation. If an equation, representing a family of curves, contains n
arbitrary constants, then we differentiate the given equation n times to obtain n more equations. Using all these
equations, we eliminate the constants. The equation so obtained is the differential equation of order n for the family
of given curves.
Consider a family of curves f(x, y, a1, a2....., an ) = 0 ......(i)
where a1, a2,......an are n independent parameters.
Equation (i) is known as an n parameter family of curves e.g. y = mx is a one-parameter family of straight lines.
x 2 + y 2 + ax + by = 0 is a two parameters family of circles.
If we differentiate equation (i) n times w.r.t. x, we will get n more relations between x, y, a1, a2,.....an and
derivatives of y w.r.t. x. By eliminating a1, a2,...., an from these n relations and equation (i), we get a differential
equation.
Clearly order of this differential equation will be n i.e. equal to the number of independent parameters in the
family of curves.
Algorithm for formation of differential equations
Step (i) : Write the given equation involving independent variable x (say), dependent variable y (say) and the
arbitrary constants.
Step (ii) : Obtain the number of arbitrary constants in step (i). Let there be n arbitrary constants.
Step (iii) : Differentiate the relation in step (i) n times with respect to x.
Step (iv) : Eliminate arbitrary constants with the help of n equations involving differential coefficients obtained
in step (iii) and an equation in step (i). The equation so obtained is the desired differential equation.
Example: 10 Differential equation whose general solution is y = c1x + c2 for all values of c1 and c2 is
Solution: (d) x
(a) d2y + x2 + dy = 0 (b) d2y + y − dy = 0 (c) d2y − 1 dy = 0 (d) d2y + 1 dy − y =0
dx 2 y dx dx 2 x2 dx dx 2 2x dx dx 2 x dx x2
y = c1x + c2 .....(i)
x
There are two arbitrary constants. To eliminate these constants, we need to differentiate (i) twice.
Differentiating (i) with respect to x,
dy = c1 − c2 .....(ii)
dx x2
Again differentiating with respect to x,
d2y = 2c2 ......(iii)
dx 2 x3
From (iii), c2 = x3 d2y and from (ii), c1 = dy + c2 ;
2 dx 2 dx x2
∴ c1 = dy + x d2y
dx 2 dx 2
From (i), y = dy + x ⋅ d2 y x + x2 d2y ⇒ y = x2 d2y + x dy
dx 2 dx 2 ⋅ dx 2 dx 2 dx
2
∴ d2y 1 dy y =0
dx2 + x dx − x2
Example: 11 y = x x is a solution of the differential equation
Solution: (b) +1
Example: 12
Solution: (a) (a) y2 dy = x2 (b) x2 dy = y2 (c) y dy = x (d) x dy = y
dx dx dx dx
Example: 13
Solution: (b) We have y= x ⇒ 1 = x +1 =1+ 1
x +1 y x x
Differentiating w.r.t. x,
− 1 dy =0− 1
y2 dx x2
∴ x2 dy = y2
dx
The differential equation of all parabolas whose axes are parallel to y axis is
(a) d3y = 0 (b) d2x = c (c) d3y + d2y = 0 (d) d2y +2 dy = c
dx 3 dy2 dx 3 dx 2 dx 2 dx
The equation of a parabola whose axis is parallel to y-axis may be expressed as
(x − α)2 = 4a(y − β ) ……..(i)
There are three arbitrary constants α, β and a.
We need to differentiate (i) 3 times
Differentiating (i) w.r.t. x, 2(x − α ) = 4a dy Y
dx
Again differentiating w.r.t. x,
2 = 4a d2y ⇒ d2y 1 OX
dx 2 dx2 = 2a (α, β)
Differentiating w.r.t. x,
d3y = 0
dx 3
The differential equation of family of curves whose tangent form an angle of π/4 with the hyperbola xy = c2 is
(a) dy = x2 + c2 (b) dy = x2 − c2 (c) dy c2 (d) None of these
dx x2 − c2 dx x2 + c2 dx = − x2
The slope of the tangent to the family of curves is m1 = dy
dx
Equation of the hyperbola is xy = c2 ⇒ y= c2
x
dy c2
∴ dx = − x2
∴ Slope of tangent to xy = c2 is m2 c2
= − x2
dy c2
m1 − m2 dx + x2 c2 c2
Now tan π = 1 + m1m2 ⇒ 1= c2 ⇒ dy 1 + x2 = 1 − x2
4 x2 dy dx
1− dx
∴ dy = x2 − c2
dx x2 + c2
Variable Separable type Differential Equation.
(1) Solution of differential equations : If we have a differential equation of order ‘n’ then by solving a
differential equation we mean to get a family of curves with n parameters whose differential equation is the given
differential equation. Solution or integral of a differential equation is a relation between the variables, not involving
the differential coefficients such that this relation and the derivatives obtained from it satisfy the given differential
equation. The solution of a differential equation is also called its primitive.
For example y = ex is a solution of the differential equation dy =y.
dx
(i) General solution : The solution which contains as many as arbitrary constants as the order of the
differential equation is called the general solution of the differential equation. For example, y = A cos x + B sin x is
the general solution of the differential equation d2y +y = 0. But y = A cos x is not the general solution as it
dx 2
contains one arbitrary constant.
(ii) Particular solution : Solution obtained by giving particular values to the arbitrary constants in the general
solution of a differential equation is called a particular solution. For example, y = 3 cos x + 2 sin x is a particular
solution of the differential equation d2y +y =0
dx 2
(2) Differential equations of first order and first degree : A differential equation of first order and first
degree involves x, y and dy . So it can be put in any one of the following forms: dy = f(x, y) or f x, y, dy = 0 or
dx dx dx
f(x, y)dx + g(x, y)dy = 0 where f(x, y) and g(x, y) are obviously the functions of x and y.
(3) Geometrical interpretation of the differential equations of first order and first degree : The
general form of a first order and first degree differential equation is f x, y, dy = 0 .....(i)
dx
We know that the direction of the tangent of a curve in Cartesian rectangular coordinates at any point is given
by dy , so the equation in (i) can be known as an equation which establishes the relationship between the
dx
coordinates of a point and the slope of the tangent i.e., dy to the integral curve at that point. Solving the
dx
differential equation given by (i) means finding those curves for which the direction of tangent at each point
coincides with the direction of the field. All the curves represented by the general solution when taken together will
give the locus of the differential equation. Since there is one arbitrary constant in the general solution of the
equation of first order, the locus of the equation can be said to be made up of single infinity of curves.
(4) Solution of first order and first degree differential equations : A first order and first degree
differential equation can be written as
f(x, y)dx + g(x, y)dy = 0
or dy = f(x, y) or dy = φ(x, y)
dx g(x, y)' dx
Where f(x, y) and g(x, y) are obviously the functions of x and y. It is not always possible to solve this type of
equations. The solution of this type of differential equations is possible only when it falls under the category of some
standard forms.
(5) Equations in variable separable form : If the differential equation of the form
f1(x)dx = f2 (y)dy .....(i)
where f1 and f2 being functions of x and y only. Then we say that the variables are separable in the differential
equation.
∫ ∫Thus, integrating both sides of (i), we get its solution as f1(x)dx = f2 (y)dy + C ,
where c is an arbitrary constant.
There is no need of introducing arbitrary constants to both sides as they can be combined together to give just
one.
(i) Differential equations of the type dy = f(x)
dx
To solve this type of differential equations we integrate both sides to obtain the general solution as discussed
following : dy = f(x) ⇔ dy = f (x) dx
dx
∫ ∫ ∫Integrating both sides, we obtain, dy = f(x)dx + C or y = f(x)dx + C .
(ii) Differential equations of the type dy = f (y)
dx
To solve this type of differential equations we integrate both sides to obtain the general solution as discussed
following : dy = f(y) ⇒ dx = 1 ⇒ dx = f 1 dy
dx dy f (y) (y)
∫ ∫ ∫Integrating both sides, we obtain, 1 dy C or x 1 dy C .
dx = f (y) + = f (y) +
(6) Equations reducible to variable separable form
(i) Differential equations of the form dy = f (ax + by + c) can be reduced to variable separable form by the
dx
substitution ax + by + c = Z
∴ a + b dy = dZ
dx dx
∴ dZ − a 1 = f(Z) ⇒ dZ = a + bf(Z).
dx b dx
This is variable separable form.
(ii) Differential equation of the form
dy = ax + by + c , where a = b = K (say)
dx Ax + By + C A B
∴ dy = K(Ax + By) + C
dx Ax + By + C
Put Ax + By = Z
∴ A + B dy = dZ , ∴ dZ − A 1 = KZ + C ⇒ dZ = A + B KZ + C
dx dx dx B Z+C dx Z+C
This is variable separable form and can be solved.
Example: 14 The solution of the differential equation (1 + x 2 ) dy = x(1 + y2) is [AISSE 1983]
Solution: (a) dx
Example: 15
(a) 2 tan−1 y = log(1 + x2) + c (b) tan−1 y = log(1 + x2) + c
Solution: (b)
Example: 16 (c) 2 tan−1 y + log(1 + x 2 ) + c = 0 (d) None of these
Solution: (a)
Separating the variables, we can re-write the given differential equation as
∫ ∫xdx dy 2x dx = 2 dy ⇒ 2 tan−1 y = loge(1 + x2) + c
1+ x2 1+ y2
1+ x2 = 1+ y2 ⇒
The solution of the differential equation dy = x2 + sin 3x is [DSSE 1981]
dx
(a) y = x3 + cos 3x +c (b) y = x3 − cos 3x +c (c) y = x3 + sin 3x + c (d) None of these
3 3 3 3 3
∫ ∫We have dy = (x2 + sin 3x)dx ⇒ dy = (x2 + sin 3x)dx ⇒ y= x3 − cos 3x + c
3 3
The solution of dy = 1 is
dx y2 + sin y
(a) x = y3 − cos y + c (b) y + cosy = x + c (c) x = y3 + cos y + c (d) None of these
3 3
Given equation may be re-written as dx = (y2 + sin y)dy
∫ ∫Integrating, dx = (y2 + sin y) dy
∴ x= y3 − cos y + c
3
Example: 17 The solution of the differential equation dy = (4 x + y + 1)2 is
Solution: (c) dx
(a) 4x – y + 1 = 2 tan(2x – 2c) (b) 4x – y – 1 = 2 tan(2x – 2c)
(c) 4x + y + 1 = 2 tan(2x + 2c) (d) None of these
Let 4x + y + 1 = z ⇒ 4 + dy = dz ⇒ dy = dz − 4
dx dx dx dx
∴ dy = (4 x + y + 1)2
dx
⇒ dz − 4 = z2 ⇒ dz = z2 + 4 ⇒ dz = dx ⇒ 1 tan−1 z = x + c ⇒ tan−1 4 x +y + 1 = 2x + 2c
dx dx z2 + 4 2 2 2
∴ 4x + y + 1 = 2tan (2x + 2c)
Example: 18 Solution of the differential equation dy = x + y +7 is
dx 2x + 2y + 3
(a) 6(x + y) + 11 log(3x + 3y + 10) = 9x + c (b) 6(x + y) – 11 log(3x + 3y + 10) = 9x + c
(c) 6(x + y) − 11 log x + y + 10 = 9 x + c (d) None of these
3
Solution: (b, c) Given equation may be re-written as dy = x + y +7
dx 2(x + y) + 3
Let x + y = z
⇒ 1 + dy = dz ⇒ dy = dz −1
dx dx dx dx
∴ dz −1 = z+7
dx 2z + 3
dz =1+ z+7 3z + 10 2z + 3 dz dx 2 (3z + 10) − 11 dz dx 2 dz 11 3dz dx
dx 2z + 3 2z + 3 3z + 10 3 3z + 10 3 3 9 3z + 10 =
∫ ∫ ∫⇒ = ⇒ = ⇒ = ⇒ −
⇒ 2 z − 11 log(3z + 10) = x + c1 ⇒ 6z − 11 log(3z + 10) = 9x + 9c1
3 9
∴ 6(x + y) − 11 log(3x + 3y + 10) = 9x + c [9c1 = c]
⇒ 6(x + y) − 11 log 3 x + y + 10 = 9 x + c ⇒ 6(x + y) − 11 log x + y + 10 = 9x + (c + 11 log 3)
3 3
∴ 6(x + y) − 11 log x +y+ 10 = 9x +k (k = c + 11 log 3)
3
Homogeneous Differential Equation.
(1) Homogeneous differential equation : A function f(x, y) is called a homogeneous function of degree n if
f(λx, λy ) = λn f(x, y) .
For example, f(x, y) = x 2 − y 2 + 3xy is a homogeneous function of degree 2, because f(λx,λy) =
λ2x2 − λ2y2 + 3λx .λy = λ2 f(x,y). A homogeneous function f(x, y) of degree n can always be written as
f (x, y) = xn f y or f(x, y) = y n f x . If a first-order first degree differential equation is expressible in the form
x y
dy f(x, y) where f(x, y) and g(x, y) are homogeneous functions of the same degree, then it is called a
dx = g(x, y)
homogeneous differential equation. Such type of equations can be reduced to variable separable form by the
substitution y = vx. The given differential equation can be written as dy = x n f(y / x) = f(y / x) = F y . If y = vx ,
dx x n g(y / x) g(y / x) x
then dy = v + x dv . Substituting the value of dy = F y , we get v + x dv = F(v) ⇒ dv = dx . On
dx dx dx x dx F(v) − v x
1 dx where C is an arbitrary constant of integration. After integration, v will be
F(v) − x
∫ ∫integration, v dv = + C
replaced by y in complete solution.
x
(2) Algorithm for solving homogeneous differential equation
Step (i) : Put the differential equation in the form dy = φ(x, y)
dx ψ (x, y)
Step (ii) : Put y = vx and dy = v + x dv in the equation in step (i) and cancel out x from the right hand side.
dx dx
The equation reduces to the form v + x dv = F(v) .
dx
Step (iii) : Shift v on RHS and separate the variables v and x
Step (iv) : Integrate both sides to obtain the solution in terms of v and x.
Step (v) : Replace v by y in the solution obtained in step (iv) to obtain the solution in terms of x and y.
x
(3) Equation reducible to homogeneous form
A first order, first degree differential equation of the form
dy = a1 x + b1y + c1 , where a1 ≠ b1 .....(i)
dx a2x + b2y + c2 a2 b2
This is non-homogeneous.
It can be reduced to homogeneous form by certain substitutions. Put x = X + h, y = Y + k
Where h and k are constants, which are to be determined.
dy = dy . dY . dX = dY
∴ dx dY dX dx dX
Substituting these values in (i), we have dY = (a1 X + b1Y) + a1h + b1k + c1 .....(ii)
dX (a2 X + b2Y) + a2h + b2k + c 2
Now h, k will be chosen such that a1h + b1k + c1 = 0 .....(iii)
a2h + b2k + c2 = 0
i.e. b1c 2 h = c1a2 k = 1 .....(iv)
− b2c1 − c2a1 a1b2 − a2b1
For these values of h and k the equation (ii) reduces to dY = a1 X + b1Y which is a homogeneous differential
dX a2 X + b2Y
equation and can be solved by the substitution Y = vX. Replacing X and Y in the solution so obtained by x − h
and y − k respectively, we can obtain the required solution in terms of x and y.
Example: 19 The solution of the differential equation x dy = y(log y − log x + 1) is [IIT 1986]
Solution: (a) dx
(d) None of these
(a) y = xe cx (b) y + xecx = 0 (c) y + ex = 0
Given equation may be expressed as dy = y log y + 1 …….(i)
dx x x
Let y =v ⇒ y = vx ⇒ dy = v + x dv
x dx dx
dv dv dv dx 1 dx
dx dx v log v x log x
∫ ∫∴ From (i),
v + x = v(log v + 1) ⇒ x = v log v ⇒ = ⇒ v d(log v) =
∴ log (log v) = log x + log c ⇒ log (log v) = log (cx) ⇒ log v = cx ⇒ v = ecx ⇒ y = ecx , ∴ y = xecx
x
Example: 20 The solution of differential equation yy′ = x y2 + φ(y2 / x2) is
Solution: (a) x2 φ ′(y 2 / x2)
Example: 21
Solution: (a) (a) φ(y2 / x2) = cx2 (b) x2φ(y2 / x2) = c2y2 (c) x2φ(y2 / x2) = c (d) φ(y2 / x2) = cy
x
Example: 22
Solution: (b) Given equation may be re-written as y ⋅ dy = y 2 + φ((y / x)2) .....(i)
x dx x φ′((y / x)2)
Let y = vx ⇒ dy = v+ x dv and y =v
dx dx x
∴ From (i), vv + x dv = v2 + φ(v2) ⇒ vx dv = φ(v2) ⇒ φ ′(v 2 )(2v dv) = 2 dx
dx φ ′(v 2 ) dx φ ′(v 2 ) φ(v2) x
Integrating, ln(φ(v2)) = 2 ln x + ln c ⇒ φ(v2) = cx2
∴ φ(y2 / x2) = cx2
The solution of dy = y3 + 2x2y is
dx x3 + 2xy2
(a) (x2 − y2)3 = Bx2y2 (b) (x2 + y2)3 = Bx2y2 (c) (x2 − y2)3 = x2y2 (d) None of these
Given equation is homogeneous. Let y = vx ∴ dy =v+x dv
dx dx
⇒ y3 + 2x2y =v+x dv ⇒ (y / x)3 + 2(y / x) = v + x dv ⇒ v3 + 2v = v+ x dv ⇒ x dv = v1v+2 +2 − 1 = v11+−2vv22
x3 + 2xy2 dx 1 + 2(y / x)2 dx 1 + 2v2 dx dx 2v2
⇒ 1 + 2v2 dv = dx ⇒ 1 + 2v2 dv = dx ⇒ A + B + 1 D dv = dx
v(1 − v2) x v(1 − v)(1 + v) x v 1−v +v x
where A(1 − v)(1 + v) + Bv(1 + v) + Dv(1 − v) = 1 + 2v2
Putting v = 0, A = 1
v = 1, B = 3
2
v = – 1, D = − 3
2
∴ 1 + 3 1 − 3 1 1 v dv = dx
v 2 1−v 2 + x
Integrating both side, we get ln v + 3 ln(1 − v) − 3 ln(1 + v) = ln x + ln c ⇒ ln v − 3 ln(1 − v) − 3 ln(1 + v) = ln cx
2 − 1 2 2 2
⇒ v /{(1 − v)1 + v}3 / 2 = cx ⇒ v 2 = (1 − v2)3 ⇒ y 2 = 1 − y2 3 ⇒ (x 2 − y2)3 = x 2y 2
cx cx 2 x2 c2
∴ (x2 − y2)3 = Bx2y2 , 1 = B
c2
The solution of dy = x − 3y + 2 is
dx 3x − y + 6
(a) y2 + 6(x + 2)y + (x + 2)2 = c (b) y2 − 6(x + 2)y + (x + 2)2 = c
(c) y2 − 6(y + 2)x + x2 = c (d) None of these
Given equation is non-homogeneous
Let x = X + h, y = Y + k
⇒ dy = dY
dx dX
∴ dY = (X + h) − 3(Y + k) + 2 = X − 3Y + (h − 3k + 2)
dX 3(X + h) − (Y + k) + 6 3X − Y + (3h − k + 6)
Let us select h and k so that h – 3k + 2 = 0 and 3h – k + 6 = 0
Solving, k = 0, h = – 2 ∴ X = x – h = x + 2, Y = y − k = y
∴ dY = X − 3Y , which is homogeneous
dX 3X − Y
Now, let Y = vX
⇒ dY =v+ X dv ⇒ X − 3Y =v+ X dv ⇒ 1 − 3(Y / X) = v + X dv ⇒ 1 − 3v =v+ X dv
dX dX 3X − Y dX 3 − (Y / X) dX 3−v dX
⇒ X dv = 1 − 3v −v = v2 − 6v + 1 ⇒ (3 − v)dv = dX ⇒ 2v − 6 1 dv = −2 dX
dX 3−v 3−v v2 − 6v + 1 X v2 − 6v + X
Integrating, ln(v2 − 6v + 1) = −2 ln X + ln c ⇒ ln(v2 − 6v + 1) + ln X 2 = ln c ⇒ X 2(v2 − 6v + 1) = c ⇒ Y 2 − 6XY + X 2 = c
∴ y2 − 6(x + 2)y + (x + 2)2 = c
Exact Differential Equation.
(1) Exact differential equation : If M and N are functions of x and y, the equation Mdx + Ndy = 0 is called
exact when there exists a function f(x, y) of x and y such that
d[f(x, y)] = Mdx + Ndy i.e., ∂f dx + ∂f dy = Mdx + Ndy
∂x ∂y
where ∂f = Partial derivative of f(x, y) with respect to x (keeping y constant)
∂x
∂f = Partial derivative of f(x, y) with respect to y (treating x as constant)
∂y
Note : An exact differential equation can always be derived from its general solution directly by
differentiation without any subsequent multiplication, elimination etc.
(2) Theorem : The necessary and sufficient condition for the differential equation Mdx + Ndy = 0 to be exact
is ∂M = ∂N i.e., partial derivative of M(x, y) w.r.t. y = Partial derivative of N(x, y) w.r.t. x
∂y ∂x
(3) Integrating factor : If an equation of the form Mdx + Ndy = 0 is not exact, it can always be made exact
by multiplying by some function of x and y. Such a multiplier is called an integrating factor.
(4) Working rule for solving an exact differential equation :
Step (i) : Compare the given equation with Mdx + Ndy = 0 and find out M and N. Then find out ∂M and
∂y
∂N . If ∂M = ∂N , the given equation is exact.
∂x ∂y ∂x
Step (ii) : Integrate M with respect to x treating y as a constant.
Step (iii) : Integrate N with respect to y treating x as constant and omit those terms which have been already
obtained by integrating M.
Step (iv) : On adding the terms obtained in steps (ii) and (iii) and equating to an arbitrary constant, we get the
required solution.
∫ ∫In other words, solution of an exact differential equation is Mdx + Ndy = c
Regarding y Only those terms
as constant not containing x
(5) Solution by inspection : If we can write the differential equation in the form
f(f1(x, y))d(f1(x, y)) + φ(f2(x, y))d(f2(x, y)) + ...... = 0 , then each term can be easily integrated separately. For this the
following results must be memorized.
(i) d(x + y) = dx + dy (ii) d(xy) = xdy + ydx
(iii) d x = ydx − xdy (iv) d y = xdy − ydx
y y2 x x2
(v) d x2 = 2xydx − x 2dy (vi) d y2 = 2xydy − y 2dx
y y2 x x2
(vii) d x 2 = 2xy 2dx − 2x 2ydy (viii) d y 2 = 2x 2ydy − 2xy 2dx
y 2 y4 x 2 x4
(ix) d tan −1 x = ydx − xdy (x) d tan −1 y = xdy − ydx
y x2 + y2 x x2 + y2
(xi) d[ln(xy)] = xdy + ydx (xii) d ln x = ydx − xdy
xy y xy
(xiii) d 1 ln(x 2 + y 2 ) = xdx + ydy (xiv) d ln y = xdy − ydx
2 x2 + y2 x xy
(xv) d − 1 = xdy + ydx (xvi) d ex = ye x dx − e x dy
xy x 2y 2 y y2
(xvii) d ey = xeydy − eydx (xviii) d (x myn ) = x m−1yn−1(mydx + nxdy)
x x2
(xix) d x2 + y2 = xdx + y dy (xx) d 1 log x + y = x dy − y dx
x2 + y2 2 x − y x2 − y2
(xxi) d [ f(x, y)]1−n = f ′(x, y)
1−n (f(x, y))n
Example: 23 The general solution of the differential equation (x + y)dx + xdy = 0 is [MP PET 1994, 95]
Solution: (c)
(a) x 2 + y 2 = c (b) 2x2 − y2 = c (c) x2 + 2xy = c (d) y2 + 2xy = c
Example: 24
Solution: (a) We have xdx + (ydx + xdy) = 0 ⇒ xdx + d(xy) = 0 (d) None of these
Integrating, x2 + xy = c
2 2
∴ x2 + 2xy = c
Solution of y(2xy + ex )dx = exdy is
(a) yx2 + ex = cy (b) xy2 + ex = cx (c) xy2 + e−x = c
Re-writing the given equation,
2xy2dx + ye xdx = exdy ⇒ 2x dx + ye xdx − exdy =0 ⇒ d(x 2 ) + d ex = 0
y2 y
Integrating, x2 + ex =c
y
∴ yx 2 + ex = cy
Example: 25 Solution of (x2 − 4 xy − 2y2)dx + (y2 − 4 xy − 2x2)dy = 0 is
Solution: (a)
(a) x3 + y3 − 6xy(x + y) = c (b) x3 + y3 + 6xy(x − y) = c
(c) x3 + y3 + 6xy(x + y) = c (d) x3 + y3 − 6xy(x − y) = c
Comparing given equation with Mdx + Ndy = 0,
We get, M = x2 − 4 xy − 2y2 , N = y2 − 4 xy − 2x2
∂M = −4 x − 4y
∂y
∂N = −4y − 4 x
∂x
∴ ∂M = ∂N
∂y ∂x
So the given differential equation is exact.
Integrating m w.r.t. x, treating y as constant,
∫ ∫Mdx = (x2 − 4 xy − 2y2)dx = x3 − 2x2y − 2y2x
3
Integrating N w.r.t. y, treating x as constant,
(y2 − 4 xy − 2x2)dy = y3 − 2xy2 − 2x2y = y3 ; (omitting − 2xy2 − 2x2y which already occur in Mdx )
∫ ∫ ∫Ndy = 3 3
∴ Solution of the given equation is x3 − 2x2y − 2xy2 + y3 =λ ⇒ x3 + y3 − 6xy(x + y) = 3λ
3 3
∴ x3 + y3 − 6xy(x + y) = c (3λ = c)
Linear Differential Equation.
(1) Linear and non-linear differential equations : A differential equation is a linear differential equation if
it is expressible in the form Po dny + P1 d n−1y + P2 d n−2y + ... + Pn−1 dy + Pn y = Q where P0 , P1, P2 ,..., Pn−1, Pn and
dx n dx n−1 dx n−2 dx
Q are either constants or functions of independent variable x .
Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and
dependent variable in the first power and there are no product of these, and also the coefficient of the various terms
are either constants or functions of the independent variable, then it is said to be linear differential equation.
Otherwise, it is a non linear differential equation.
It follows from the above definition that a differential equation will be non-linear differential equation if (i) its
degree is more than one (ii) any of the differential coefficient has exponent more than one. (iii) exponent of the
dependent variable is more than one. (iv) products containing dependent variable and its differential coefficients are
present.
(2) Linear differential equation of first order : The general form of a linear differential equation of first
order is
dy + Py = Q .....(i)
dx
Where P and Q are functions of x (or constants)
For example, dy + xy = x 3, x dy + 2y = x 3 , dy + 2y = sin x etc. are linear differential equations. This type of
dx dx dx
differential equations are solved when they are multiplied by a factor, which is called integrating factor, because by
multiplication of this factor the left hand side of the differential equation (i) becomes exact differential of some
function.
Multiplying both sides of (i) by e ∫ Pdx , we get e ∫ Pdx dy + Py = Q e ∫ Pdx ⇒ d y e ∫ Pdx = Q e ∫ Pdx
dx dx
On integrating both sides w. r. t. x, we get ; ∫y e ∫ Pdx = Q e ∫ Pdx dx + C ……..(ii)
which is the required solution, where C is the constant of integration. e ∫ Pdx is called the integrating factor. The
∫solution (ii) in short may also be written as y.(I.F.) = Q.(I.F.) dx + C
(3) Algorithm for solving a linear differential equation :
Step (i) : Write the differential equation in the form dy + Py = Q and obtain P and Q.
dx
Step (ii) : Find integrating factor (I.F.) given by I.F. = e ∫ Pdx .
Step (iii) : Multiply both sides of equation in step (i) by I.F.
∫Step (iv) : Integrate both sides of the equation obtained in step (iii) w. r. t. x to obtain y(I.F.) = Q(I.F.) dx + C
This gives the required solution.
(4) Linear differential equations of the form dx + Rx = S . Sometimes a linear differential equation can be
dy
put in the form dx + Rx = S where R and S are functions of y or constants. Note that y is independent variable
dy
and x is a dependent variable.
(5) Algorithm for solving linear differential equations of the form dx + Rx = S
dy
Step (i) : Write the differential equation in the form dx + Rx =S and obtain R and S.
dy
Step (ii) : Find I.F. by using I.F. = e ∫ R dy
Step (iii) : Multiply both sides of the differential equation in step (i) by I.F.
Step (iv) : Integrate both sides of the equation obtained in step (iii) w. r. t. y to obtain the solution given by
∫x(I.F.) = S (I.F.) dy + C where C is the constant of integration.
(6) Equations reducible to linear form (Bernoulli's differential equation) : The differential equation of
type dy + Py = Qy n ......(i)
dx
Where P and Q are constants or functions of x alone and n is a constant other than zero or unity, can be
reduced to the linear form by dividing by yn and then putting y −n+1 = v , as explained below.
Dividing both sides of (i) by yn , we get y −n dy + Py −n+1 = Q
dx
Putting y −n+1 =v so that (−n + 1)y −n dy = dv , we get 1 dv + Pv = Q ⇒ dv + (1 − n)Pv = (1 − n)Q which is
dx dx − n + 1 dx dx
a linear differential equation.
Remark : If n = 1 , then we find that the variables in equation (i) are separable and it can be easily integrated
by the method discussed in variable separable from.
(7) Differential equation of the form : dy + Pφ(y) = Qψ (y)
dx
where P and Q are functions of x alone or constants.
Dividing by ψ(y), we get 1 dy + φ(y) P = Q
ψ (y) dx ψ (y)
Now put φ(y) = v , so that d φ(y) = dv or dv = k ⋅ 1 dy , where k is constant
ψ (y) dx ψ (y) dx dx ψ (y) dx
We get dv + kP v = kQ
dx
Which is linear differential equation.
Example: 26 Which of the following is a linear differential equation
Solution: (c)
Example: 27 (a) d2y 2 + x 2 dy 2 = 0 (b) y = dy + 1 + dy 2 (c) dy + y = log x (d) y dy − 4 = x
Solution: (a) dx 2 dx dx dx dx x dx
Example: 28 (a), (b), (d) do not fulfill the criteria of a linear differential equation but (c) do.
Solution: (b)
dy + y = log x is a linear differential equation.
dx x
Find the integral factor of equation (x2 + 1) dy + 2xy = x2 −1 [UPSEAT 2002]
dx
(d) None of these
(a) x2 + 1 (b) 2x (c) x2 −1
x2 +1 x2 +1
Given equation may be written as dy + 2x y = x2 −1
dx x2 +1 x2 +1
Comparing with dy + Py = Q ,
dx
P = 2x
x2 +1
Pdx 2 xdx
=e 1+ x 2 = eln(1+ x 2 ) = 1 + x 2
∫ ∫I.F. = e
The solution of dy + 2y tan x = sin x is
dx
(a) y sec3 x = sec2 x + c (b) y sec2 x = sec x + c (c) y sin x = tan x + c (d) None of these
Comparing with dy + Py = Q ,
dx
P = 2 tan x , Q = sin x
∫I.F. = e 2 tan xdx = e2 ln sec x = eln sec 2 x = sec 2 x
∫ ∫Multiplying given equation by I.F. and integrating, y sec2 x = sin x. sec2 xdx = sec x tan xdx
∴ y sec2 x = sec x + c
Example: 29 The solution of the differential equation (1 + y 2 ) + (x − e tan −1 y ) dy =0 is [AIEEE 2003]
Solution: (b) dx
Example: 30 (a) (x − 2) = ketan−1 y (b) 2xetan−1 y = e2 tan−1 y + k
Solution: (b)
(c) xetan−1 y = tan−1 y + k (d) xe2 tan−1 y = etan−1 y + k
Example: 31
Solution: (d) We have (x e tan −1 y ) dy = −(1 + y2) ⇒ dx x − e tan−1 y dx 1 x e tan−1 y ……..(i)
dx dy 1+ y2 dy + 1 + y2 1+ y2
− = − ⇒ =
This is a linear differential equation of the form dx + R(y).x = S(y)
dy
R = 1 , S = e tan−1 y
1+ y2 1+ y2
Integrating factor = e∫ Rdy = e∫ dy = e tan−1 y
1+ y 2
e tan−1 y (etan−1 y )2 dy (etan−1 y )2 k
1+ y2 1+ y2 = 2 + 2
∫ ∫Multiplying (i) by I.F. and integrating, xetan−1 y =
⋅ etan−1 ydy =
∴ 2xetan−1 y = e2 tan−1 y + k
Solution of dy − y tan x = −y2 sec x is
dx
(a) y sec x = tan x + c (b) sec x = tan x +c (c) y cos x = tan x + c (d) None of these
y
Re-writing the given equation, y−2 dy − y−1 tan x = − sec x
dx
Let y−1 =v ⇒ − y−2 dy = dv
dx dx
∴ dv + tan x .v = sec x ……..(i)
dx
∫I.F. = e tan x = eln sec x = sec x
∫Multiplying (i) by sec x and integrating, v sec x = sec2 xdx = tan x + c
∴ sec x = tan x + c
y
The solution of dz z log z = z (log z)2 is
dx + x x2
(a) 1 z x = 2 − x 2c (b) 1 z x = 2 + x 2c (c) 1 z x = x 2c (d) 1 z x = 1 + cx 2
log log log log 2
Dividing the given equation by z(log z)2 , 1 dz 1 1 1
z(log z)2 dx + x log z = x2
Let 1 =t 1 1 dz dt
log z ⇒ − (log z)2 ⋅ z dx = dx
∴ − dt + t = 1
dx x x2
⇒ dt t 1 ……..(i)
dx − x = − x2
∫I.F. = e dx = e−ln x = eln 1 = 1
−x x x
1 and integrating, t 1 1 1 1
∫Multiplying (i) by x x = − x3 dx = 2x2 +c ⇒ x log z = 2x2 +c
∴ 1 x = 1 + cx 2
ln z 2
Application of Differential Equation.
Differential equation is applied in various practical fields of life. It is used to define various physical laws and
quantities. It is widely used in physics, chemistry, engineering etc.
Some important fields of application are ;
(i) Rate of change (ii) Geometrical problems etc.
Differential equation is used for finding the family of curves for which some conditions involving the derivatives
are given.
Equation of the tangent at a point (x, y) to the curve y = f(x) is given by Y − y = dy (X − x) ……..(i)
dx
and equation of normal at (x, y) is Y −y =− 1 (X − x) ……..(ii)
dy
dx
y 0, y dy
dy 0 dx
The tangent meets X-axis at x − dx , and Y-axis at − x
x dy 0 x
dx
The normal meets X-axis at + y , and Y-axis at 0, y +
dy
dx
Example: 32 A particle moves in a straight line with a velocity given by dx = (x + 1) (x is the distance described). The time taken by a
Solution: (b) dt
Example: 33
particle to transverse a distance of 99 metres
(a) log10 e (b) 2 loge 10 (c) log10 e (d) 1 log10 e
2
We have dx = dt
x +1
99 dx t [ln(x + 1)]099
0 x +1 =
∫ ∫Integrating, dt ⇒ =t
0
∴ t = ln 100 = loge(10)2 = 2 loge 10
The slope of the tangent at (x, y) to a curve passing through 1, π as given by y − cos 2 y , then the equation of the
4 x x
curve is [Kurukshetra CEE 2002]
(a) y = tan−1 log e (b) y = x tan−1 log x (c) y = x tan−1 log e (d) None of these
x e x
Solution: (c) We have dy = y − cos 2 y
dx x x
Example: 34
Solution: (d) Let y = vx ⇒ dy = v+ x dv ⇒ v+ x dv = v − cos2 v ⇒ x dv = − cos2 v ⇒ sec 2 vdv = − dx ⇒ tan v = – ln x + c
dx dx dx dx x
⇒ tan (y/x) = – ln x + c
For x = 1, y = π/4
⇒ tan π / 4 = − ln 1 + c ⇒ 1 = 0 + c
∴c=1
∴ tan(y / x) = 1 − ln x
⇒ y/ x = tan−1(1 − ln x) = tan−1(ln e − ln x) = tan−1 ln e
x
∴ y= x tan−1 ln e
x
The equation of the curve which is such that the portion of the axis of x cut off between the origin and tangent at any point
is proportional to the ordinate of that point (b is constant of proportionality)
(a) y = (a − x x) (b) log x = by2 + a (c) x2 = y(a − b log y) (d) None of these
b log
Tangent at P(x, y) to the curve y = f(x) may be expressed as Y −y = dy (X − x)
dx
∴ Q = x − y dx , 0 P(x, y)
dy
As per question, OQ ∝ y
⇒ x − y dx ∝ y ⇒ x −y dx = by ⇒ x − dx = b (0, 0) O Q
dy dy y dy
∴ dx = x − b
dy y
Let x =v ⇒ x = vy ⇒ dx = v+ y dv ⇒ x − b = v + y dv ⇒ v − b = v + y dv ⇒ − b = y dv ⇒ −b dy = dv
y dy dy y dy dy dy y
∫ ∫Integrating, dy x (a, an arbitrary constant)
dv = −b y ⇒ v = −b ln y + a ⇒ y = a − b ln y
∴ x = y(a − b ln y)
Miscellaneous Differential Equation.
(1) A special type of second order differential equation : d2y = f(x) ……..(i)
dx 2 ……..(ii)
Equation (i) may be re-written as d dy = f(x) ⇒ d dy = f (x)dx
dx dx dx
∫Integrating, dy = f(x)dx + c1 i.e. dy = F(x) + c1
dx dx
∫Where F(x) = f(x)dx + c1dx
From (ii), dy = f(x)dx + c1dx
∫Integrating, y = F(x)dx + c1 x + c2
∴ y = H(x) + c1x + c2
∫where H(x) = F(x)dx c1 and c2 are arbitrary constants.
(2) Particular solution type problems : To solve such a problem, we proceed according to the type of the
problem (i.e. variable-separable, linear, exact, homogeneous etc.) and then we apply the given conditions to find
the particular values of the arbitrary constants.
Example: 35 The solution of the equation x2 d2y = ln x when x = 1, y = 0 and dy = −1 is [Orissa JEE 2003]
Solution: (d) dx 2 dx
Example: 36 (a) 1 (ln x)2 + ln x (b) 1 (ln x)2 − ln x (c) − 1 (ln x)2 + ln x (d) − 1 (ln x)2 − ln x
Solution: (d) 2 2 2 2
We have d2y = ln x ⇒ d dy = ln x dx
dx 2 x2 dx x2
∫ ∫Integrating, dy = ln xd − 1 = − ln x + 1 dx = − ln x − 1 + c ⇒ dy = − 1 + ln x + c
dx x x x2 x x dx x
When x = 1, dy = −1
dx
∴–1=–1+c⇒c=0
dy 1 + ln x dy 1 + ln x dx dy = + dx ln x. 1 dx y ln x 1 (ln x)2
dx x x x+ x 2
∫ ∫ ∫∴ = − ⇒ = − ⇒ − ⇒ − = + + λ
y = 0 when x = 1
∴ 0 = 0 + 02 + λ ⇒ λ =0 ⇒ − y = ln x + 1 (ln x)2
2
∴ y = − 1 (ln x)2 − ln x
2
A continuously differentiable function φ(x) in (0, π) satisfying y′ = 1 + y2 , y(0) = 0 = y(π ) is
(a) tan x (b) x(x – π) (c) (x − π )(1 − ex ) (d) Not possible
y′ = 1 + y2 dy y2 dy dx ⇒ tan−1 y = x + c
dx 1+ y2 =
∫ ∫For φ(x) = y,
⇒ =1+ ⇒
∴ y = tan (x + c)
i.e., φ(x) = tan (x + c)
As y(0) = 0, 0 = tan c and y(π) = 0 ⇒ 0 = tan (π + c) = tan c
∴c=0
∴ φ(x) = y = tan x.
But tan x is not continuous in (0, π)
Since tan π is not defined.
2
Hence there exists not a function satisfying the given condition.
Linear programing
‘Linear Programming’ is a scientific tool to handle optimization problems. Here, we shall learn about some
basic concepts of linear programming problems in two variables, their applications, advantages, limitations,
formulation and graphical method of solution.
Linear Inequations .
(1) Graph of linear inequations
(i) Linear inequation in one variable : ax + b > 0, ax + b < 0, cy + d > 0 etc. are called linear inequations in
one variable. Graph of these inequations can be drawn as follows :
Y x = − b
a
X′ O
ax+b<0 X
Y′ ax+b>0
The graph of ax + b > 0 and ax + b < 0 are obtained by dividing xy-plane in two semi-planes by the line
x = b (which is parallel to y-axis). Similarly for cy +d > 0 and cy + d < 0 .
−a
Y
cy+d>0 y = − d
c
X′ X
cy+d<0
Y′
(ii) Linear Inequation in two variables : General form of these inequations are ax + by > c,ax + by < c . If
any ordered pair (x1, y1 ) satisfies some inequations, then it is said to be a solution of the inequations.
The graph of these inequations is given below (for c > 0) :
Y
ax+by<c ax+by=c
X′ O X
ax+by>c
Working rule Y′
To draw the graph of an inequation, following procedure is followed :
(i) Write the equation ax + by = c in place of ax + by < c and ax + by > c .
(ii) Make a table for the solutions of ax + by = c .
(iii) Now draw a line with the help of these points. This is the graph of the line ax + by = c .
(iv) If the inequation is > or <, then the points lying on this line is not considered and line is drawn dotted or
discontinuous.
(v) If the ineuqation is > or <, then the points lying on the line is considered and line is drawn bold or
continuous.
(vi) This line divides the plane XOY in two region.
To Find the region that satisfies the inequation, we apply the following rules:
(a) Take an arbitrary point which will be in either region.
(b) If it satisfies the given inequation, then the required region will be the region in which the arbitrary point is located.
(c) If it does not satisfy the inequation, then the other region is the required region.
(d) Draw the lines in the required region or make it shaded.
(2) Simultaneous linear inequations in two variables : Since the solution set of a system of simultaneous
linear inequations is the set of all points in two dimensional space which satisfy all the inequations simultaneously.
Therefore to find the solution set we find the region of the plane common to all the portions comprising the solution
set of given inequations. In case there is no region common to all the solutions of the given inequations, we say that
the solution set is void or empty.
(3) Feasible region : The limited (bounded) region of the graph made by two inequations is called feasible
region. All the points in feasible region constitute the solution of a system of inequations. The feasible solution of a
L.P.P. belongs to only first quadrant. If feasible region is empty then there is no solution for the problem.
Example: 1 Inequations 3x − y ≥ 3 and 4 x − y > 4
Solution : (a)
(a) Have solution for positive x and y (b) Have no solution for positive x and y
Example: 2
Solution : (b) (c) Have solution for all x (d) Have solution for all y
Following figure will be obtained on drawing the graphs of given inequations :
From 3x −y ≥ 3, x + y = 1 Y
1 −3
From 4x −y ≥ 4, x + y =1 O (1,0) X
1 −4 [MP PET 1999]
Clearly the common region of both the inequations is true for positive value (0,–3)
of (x, y). It is also true for positive values of x and negative values of y.
The constraints (0,–4)
– x1 + x2 + ≤ 1 ; – x1 + 3x2 ≤ 9 ; x, x 2 ≥ 0 define
(a) Bounded feasible space (b) Unbounded feasible space
(c) Both bounded and unbounded feasible space (d) None of these
x2 (3,4)
(0,3)
(–9,0) (0,1) x1
O (–1,0)
It is clear from the graph, the constraints define an unbounded feasible space.
Example: 3 Shaded region is represented by Y [MP PET 1997]
(0,20) x+y=2
C(0,16) B 20 , 40
3 3
y Shaded 2x+5y=80
region
A(20,0) X
Ox (40,0)
(a) 2x + 5y ≥ 80, x + y ≤ 20, x ≥ 0, y ≤ 0 (b) 2x + 5y ≥ 80, x + y ≥ 20, x ≥ 0, y ≥ 0
(c) 2x + 5y ≤ 80, x + y ≤ 20, x ≥ 0, y ≥ 0 (d) 2x + 5y ≤ 80, x + y ≤ 20, x ≤ 0, y ≤ 0
Solution: (c) In all the given equations, the origin is present in shaded area. Answer (c) satisfy this condition.
Example: 4
Shaded region is represented by
Shaded 4x–2y=–3
region A(0,3/2)
y
O x
B −3 ,0
4
(a) 4 x − 2y ≤ 3 (b) 4 x − 2y ≤ −3 (c) 4 x − 2y ≥ 3 (d) 4 x − 2y ≥ −3
Solution: (b) Origin is not present in given shaded area. So 4 x − 2y ≤ −3 satisfy this condition.
Terms of Linear Programming .
The term programming means planning and refers to a process of determining a particular program.
(1) Objective function : The linear function which is to be optimized (maximized or minimized) is called
objective function of the L.P.P.
(2) Constraints or Restrictions : The conditions of the problem expressed as simultaneous equations or
inequalities are called constraints or restrictions.
(3) Non-negative Constraints : Variables applied in the objective function of a linear programming problem are
always non-negative. The ineqaulities which represent such constraints are called non-negative constraints.
(4) Basic variables : The m variables associated with columns of the m × n non-singular matrix which may
be different from zero, are called basic variables.
(5) Basic solution : A solution in which the vectors associated to m variables are linear and the remaining
(n − m) variables are zero, is called a basic solution. A basic solution is called a degenerate basic solution, if at least
one of the basic variables is zero and basic solution is called non-degenerate, if none of the basic variables is zero.
(6) Feasible solution : The set of values of the variables which satisfies the set of constraints of linear
programming problem (L.P.P) is called a feasible solution of the L.P.P.
(7) Optimal solution : A feasible solution for which the objective function is minimum or maximum is called
optimal solution.
(8) Iso-profit line : The line drawn in geometrical area of feasible region of L.P.P. for which the objective
function remains constant at all the points lying on the line, is called iso-profit line.
If the objective function is to be minimized then these lines are called iso-cost lines.
(9) Convex set : In linear programming problems feasible solution is generally a polygon in first quadrant .
This polygon is convex. It means if two points of polygon are connecting by a line, then the line must be inside to
polygon. For example,\
AA
BB
(i) (ii) (iii) (iv)
Figure (i) and (ii) are convex set while (iii) and (iv) are not convex set
Mathemiatical formulation of a Linear programming problem .
There are mainly four steps in the mathematical formulation of a linear programming problem, as
mathematical model. We will discuss formulation of those problems which involve only two variables.
(1) Identify the decision variables and assign symbols x and y to them. These decision variables are those
quantities whose values we wish to determine.
(2) Identify the set of constraints and express them as linear equations/inequations in terms of the decision
variables. These constraints are the given conditions.
(3) Identify the objective function and express it as a linear function of decision variables. It might take the
form of maximizing profit or production or minimizing cost.
(4) Add the non-negativity restrictions on the decision variables, as in the physical problems, negative values
of decision variables have no valid interpretation.
Graphical solution of two variable Linear programming problem .
There are two techniques of solving an L.P.P. by graphical method. These are :
(1) Corner point method, and (2) Iso-profit or Iso-cost method.
(1) Corner point method
Working rule
(i) Formulate mathematically the L.P.P.
(ii) Draw graph for every constraint.
(iii) Find the feasible solution region.
(iv) Find the coordinates of the vertices of feasible solution region.
(v) Calculate the value of objective function at these vertices.
(vi) Optimal value (minimum or maximum) is the required solution.
Note : If there is no possibility to determine the point at which the suitable solution found, then the solution
of problem is unbounded.
If feasible region is empty, then there is no solution for the problem.
Nearer to the origin, the objective function is minimum and that of further from the origin, the
objective function is maximum.
(2) Iso-profit or Iso-cost method : Various steps of the method are as follows :
(i) Find the feasible region of the L.P.P.
(ii) Assign a constant value Z1 to Z and draw the corresponding line of the objective function.
(iii) Assign another value Z2 to Z and draw the corresponding line of the objective function.
(iv) If Z1 < Z2,(Z1 > Z2), then in case of maximization (minimization) move the line P1Q1 corresponding to Z1
to the line P2Q2 corresponding to Z2 parallel to itself as far as possible, until the farthest point within the feasible region is
touched by this line. The coordinates of the point give maximum (minimum) value of the objective function.
Note : The problem with more equations/inequations can be handled easily by this method.
In case of unbounded region, it either finds an optimal solution or declares an unbounded solution.
Unbounded solutions are not considered optimal solution. In real world problems, unlimited profit
or loss is not possible.
To find the Vertices of Simple feasible region without Drawing a Graph .
(1) Bounded region: The region surrounded by the inequalities ax + by ≤ m and cx + dy ≤ n in first
quadrant is called bounded region. It is of the form of triangle or quadrilateral. Change these inequalities into
equation, then by putting x = 0 and y = 0, we get the solution also by solving the equation in which there may be
the vertices of bounded region.
The maximum value of objective function lies at one vertex in limited region.
(2) Unbounded region : The region surrounded by the inequations ax + by ≥ m and cx + dy ≥ n in first
quadrant, is called unbounded region.
Change the inequation in equations and solve for x = 0 and y = 0 . Thus we get the vertices of feasible region.
The minimum value of objective function lies at one vertex in unbounded region but there is no existence of
maximum value.
Problems having Infeasible Solutions.
In some of the linear programming problems, constraints are inconsistent i.e. there does not exist any point
which satisfies all the constraints. Such type of linear programming
problems are said to have infeasible solution. 10
9
For Example : Maximize Z = 5x + 2y 8
subject to the constraints 7
x + y ≤ 2 , 3x + 3y ≥ 2 , x, y ≥ 0 6
The above problem is illustrated graphically in the fig. 5
There is no point satisfying the set of above constraints. Thus, the 4
problem is having an infeasible solution. 3
2
1
1 2 3 4 5 6 7 8 9 10
Some important points about L.P.P..
(1) If the constraints in a linear programming problem are changed, the problem is to be re-evaluated.
(2) The optimal value of the objective function is attained at the point, given by corner points of the feasible region.
(3) If a L.P.P. admits two optimal solutions, it has an infinite number of optimal solutions.
(4) If there is no possibility to determine the point at which the suitable solution can be found, then the
solution of problem is unbounded.
(5) The maximum value of objective function lies at one vertex in limited region.
Example: 5 A firm makes pants and shirt. A shirt takes 2 hour on machine and 3 hour of man labour while a pant takes 3 hour on
machine and 2 hour of man labour. In a week there are 70 hour of machine and 75 hour of man labour available. If the
firm determines to make x shirts and y pants per week, then for this the linear constraints are
(a) x ≥ 0, y ≥ 0,2x + 3y ≥ 70,3x + 2y ≥ 75 (b) x ≥ 0, y ≥ 0,2x + 3y ≤ 70,3x + 2y ≥ 75
(c) x ≥ 0, y ≥ 0,2x + 3y ≥ 70,3x + 2y+ ≤ 75 (d) x ≥ 0, y ≥ 0,2x + 3y ≤ 70,3x + 2y ≤ 75
Solution: (d) Shirt (x) Working time on machine Man labour
Pant (y) 2 hours 3 hours
Example: 6 Availability 3 hours 2 hours
Solution: (b) 70 hours 75 hours
Linear constraints are 2x + 3y ≤ 70,3x + 2y ≤ 75 and x, y ≥ 0
The minimum value of the objective function Z = 2x + 10y for linear constraints
x − y ≥ 0, x − 5y ≤ −5 and x, y ≥ 0 is
(a) 10 (b) 15 (c) 12 (d) 8
Required region is unbounded whose vertex is 5 , 5 Y x – y=0
4 4 x – 5y = –5
Hence the minimum value of objective function is = 2× 5 + 10 × 5 = 15 . O X
4 4
nm
∑ ∑Minimize z =
Example: 7 cij xij [MP PET 1999]
Solution: (a)
j=1 i=1
Example: 8
Solution: (d) nm
∑ ∑Subject to : xij ≤ ai ,i = 1........, m ; xij = bj, j = 1......, n is a LPP with number of constraints
j=1 i=1
(a) m + n (b) m − n (c) mn (d) m
n
(I) Condition, i = 1, x11 + x12 + x13 + ........... + x1n ≤ a1
i = 2, x 21 + x 22 + x 23 + ........... + x 2n ≤ a2
i = 3, x 31 + x 32 + x 33 + ............ + x 3n ≤ a3
...................................
i = m, xm1 + xm2 + xm3 + ........... + xmn ≤ am → m constraints
(II) Condition j = 1, x11 + x 21 + x 31 + .......... + xm1 = b1
j = 2, x12 + x 22 + x 32 + ........... + xm2 = b2
....................................
j = n, x1n + x 2n + x 3n + .............. + xmn = bn → n constraints
∴ Total constraints = m + n .
Maximize z = 3x + 2y, subject to x + y ≥ 1, y − 5x ≤ 0, x − y ≥ −1, x + y ≤ 6, x ≤ 3 and x, y ≥ 0
(a) x = 3 (b) y = 3 (c) z = 15 (d) All the above
The shaded region represents the bounded region (3,3) satisfies, so x = 3, y = 3 and z = 15 .
(0,1) y – 5x=0 x – y = –1
x=3
(5/2, 7/2)
(3,3)
x+y=6
(1,0)
Example: 9 Maximum value of 4 x + 5y subject to the constraints x + y ≤ 20, x + 2y ≤ 35, x − 3y ≤ 12 is [Kurukshetra CEE 1996]
(a) 84 (b) 95 (c) 100 (d) 96
Solution: (b) Obviously, max. 4 x + 5y = 95 . It is at (5, 15).
(0,20) x – 3y = 12
1 (5,15) (18,2)
2
0,17
(12,0)(20,0) (35,0) x + 2y = 35
x+ y = 20
Example: 10 The maximum value of Z = 4 x + 3y subjected to the constraints 3x + 2y ≥ 160, 5x + 2y ≥ 200, x + 2y ≥ 80, x, y ≥ 0 is
Solution: (d)
(a) 320 (b) 300 (c) 230 [MP PET 1998]
(d) None of these
Obviously, it is unbounded. Therefore its maximum value does not exist.
3x+2y=160
5x+2y=200 x+2y=80
Example: 11 The objective function Z = 4 x + 3y can be maximized subjected to the constraints
Solution: (c)
3x + 4y ≤ 24,8x + 6y ≤ 48, x ≤ 5, y ≤ 6; x, y ≥ 0
(a) At only one point (b) At two points only (c) At an infinite number of points (d) None of these
Obviously, the optimal solution is found on the line which is parallel to y=6
'isoprofit line'.
Hence it has infinite number of solutions.
4x+3y=0 x=5
8x+6y=48 3x+4y=24
Isoprofit line
Example: 12 The maximum value of µ = 3x + 4y subjected to the conditions x + y ≤ 40, x + 2y ≤ 60; x, y ≥ 0 is [MP PET 2002,04 ]
Solution: (d)
(a) 130 (b) 120 (c) 40 (d) 140
Obviously Max µ = 3x + 4y at (20,20) Y
µ = 60 + 80 = 140 .
A(0,30) (0,40) X
O x+y=40
B(20, 20)
x+2y=60
(60,0)
C (40, 0)
Advantages and Limitations of L.P.P..
(1) Advantages : Linear programming is used to minimize the cost of production for maximum output. In
short, with the help of linear programming models, a decision maker can most efficiently and effectively employ his
production factor and limited resources to get maximum profit at minimum cost.
(2) Limitations : (i) The linear programming can be applied only when the objective function and all the
constraints can be expressed in terms of linear equations/inequations.
(ii) Linear programming techniques provide solutions only when all the elements related to a problem can be quantified.
(iii) The coefficients in the objective function and in the constraints must be known with certainty and should
remain unchanged during the period of study.
(iv) Linear programming technique may give fractional valued answer which is not desirable in some problems.
Circle and system of circles
Definition.
A circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point in
that plane always remains the same i.e. constant. (Moving point)
The fixed point is called the centre of the circle and the fixed distance is P
called the radius of the circle. Q OR
Fixed point
Note : If r(r > 0) is the radius of a circle, the diameter d = 2r is the
Plane
maximum distance between any two points on the given circle.
• The length of the curve or perimeter (also called circumference) of circle = 2πr or πd .
• The area of circle = πr 2 or πd 2 .
4
• Line joining any two points of a circle is called chord of circle.
• Curved section between any two points of a circle is called arc of circle.
• Angle subtended at the centre of a circle by any arc = arc/radius.
• Angle subtended at the centre of a circle by an arc is double of angle subtended at the
circumference of a circle.
Standard forms of Equation of a Circle.
(1) General equation of a circle : The general equation of a circle is x 2 + y 2 + 2gx + 2 fy + c = 0,
where g, f, c are constant.
(i) Centre of the circle is (–g, –f). i.e., ( − 1 coefficient of x, − 1 coefficient of y)
2 2
(ii) Radius of the circle is g 2 + f 2 − c .
Note : The general equation of second degree ax 2 + by2 + 2hxy + 2gx + 2 fy + c = 0 represents a circle if
a = b ≠ 0 and h = 0 .
Locus of a point P represent a circle if its distance from two points A and B is not equal i.e.
PA = kPB represent a circle if k ≠ 1 .
Discussion on nature of the circle :
If g 2 + f 2 − c > 0 , then the radius of the circle will be real. Hence, in this case, it is possible to
draw a circle on a plane.
If g 2 + f 2 − c = 0 , then the radius of the circle will be zero. Such a circle is known as point circle.
If g 2 + f 2 − c < 0 , then the radius g 2 + f 2 − c of the circle will be an imaginary number.
Hence, in this case, it is not possible to draw a circle.
Special features of the general equation x 2 + y2 + 2gx + 2 fy + c = 0 of the circle :
This equation has the following peculiarities :
It is a quadratic equation in x and y.
Here the co-efficient of x 2 = the co-efficient of y2
In working out problems it is advisable to keep the co-efficient of x 2 and y2 as unity.
There is no term containing xy, i.e. the co-efficient of the term xy is zero.
This equation contains three arbitrary constants. If we want to find the equation of a circle of which
neither the centre nor the radius is known, we take the equation in the above form and determine
the values of the constants g, f, c for the circle in question from the given geometrical conditions.
Keeping in mind the above special features, we can say that the equation
ax 2 + ay 2 + 2gx + 2 fy + c = 0 …..(i) also represents a circle.
This equation can also be written as x2 + y2 + 2 g x + 2 f y + c = 0, dividing by a ≠ 0.
a a a
Hence, the centre = −g , −f and radius = g2 f2 c
a a a2 + a2 − a
(2) Central form of equation of a circle : The equation of a circle having centre (h, k) and radius r is
(x − h)2 + (y − k)2 = r 2
Note : If the centre is origin, then the equation of the circle is r P(x,y)
(h,k)
x2 + y2 = r2
C
If r = 0, then circle is called point circle and its equation is (x − h)2 + (y − k)2 = 0
(3) Concentric circle : Two circles having the same centre C (h, k) but different radii r1 and r2 respectively
are called concentric circles. Thus the circles (x − h)2 + (y − k)2 = r12 and (x − h)2 + (y − k)2 = r22 , r1 ≠ r2 are
concentric circles. Therefore, the equations of concentric circles differ only in constant terms.
(4) Circle on a given diameter : The equation of the circle drawn on the straight line joining two given
.points (x1, y1) and (x 2, y2 ) as diameter is (x − x1)(x − x 2 ) + (y − y1)(y − y2 ) = 0 P(x,y)
Note : If the coordinates of the end points of a diameter of a circle are 90°
given, we can also find the equation of the circle by finding the (x1,y1)A r B(x2,y2)
coordinates of the centre and radius. The centre is the mid-point C(h,k)
of the diameter and radius is half of the length of the diameter.
(5) Parametric coordinates
(i) The parametric coordinates of any point on the circle (x − h)2 + (y − k)2 = r 2 are given by
(h + r cosθ ,k + r sinθ ) , (0 ≤ θ < 2π )
In particular, co-ordinates of any point on the circle x 2 + y 2 = r 2 are (r cosθ ,r sinθ ) , (0 ≤ θ < 2π )
(ii) The parametric co-ordinates of any point on the circle x 2 + y 2 + 2gx + 2 fy + c = 0 are
x = −g + (g 2 + f 2 − c ) cosθ and y = − f + (g 2 + f 2 − c) sinθ , (0 ≤ θ < 2π )
(6) Equation of a circle under given conditions: The general equation of circle, i.e.,
x 2 + y 2 + 2gx + 2 fy + c = 0 contains three independent constants g, f and c. Hence for determining the equation of
a circle, three conditions are required.
(i) The equation of the circle through three non-collinear points A(x1, y1), B(x 2, y2 ), C (x3 , y3 ):
Let the equation of circle be x 2 + y 2 + 2gx + 2 fy + c = 0 …..(i)
If three points (x1, y1), (x 2, y2 ), (x3 , y3 ) lie on the circle (i), their co-ordinates must satisfy its equation. Hence
solving equations x12 + y12 + 2gx1 + 2 fy1 + c = 0 …..(ii)
x 2 + y 2 + 2gx 2 + 2 fy2 + c = 0 …..(iii)
2 2
x 2 + y 2 + 2gx 3 + 2 fy3 + c = 0 …..(iv)
3 3
g, f, c are obtained from (ii), (iii) and (iv). Then to find the circle (i).
Alternative method
(1) The equation of the circle through three non-collinear points A(x1, y1), B(x 2, y2 ), C (x3 , y3 ) is
x2 + y2 x y 1
x12 + y12 x1 y1 1 =0
x2 y2 1
x 2 + y 2
2 2
x 2 + y 2 x3 y3 1
3 3
(2) From given three points taking any two as extremities of diameter of a circle S = 0 and equation of
straight line passing through these two points is L = 0. Then required equation of circle is S + λL = 0 , where λ is a
parameter which can be found out by putting third point in the equation.
Note : Cyclic quadrilateral : If all the four vertices of a quadrilateral lie on a circle, then the
quadrilateral is called a cyclic quadrilateral. The four vertices are said to be concylic.
Equation of a Circle in Some special cases.
(1) If centre of the circle is (h,k) and it passes through origin then its equation is
(x − h)2 + (y − k)2 = h2 + k 2 ⇒ x 2 + y 2 − 2hx − 2ky = 0
(2) If the circle touches x axis then its equation is (Four cases) (x ± h)2 + (y ± k)2 = k 2
Y
(–h,k) (h,k) X
k
k
X′ k
(h,–k)
k
(–h,–k)
Y′
(3) If the circle touches y axis then its equation is (Four cases) (x ± h)2 + (y ± k)2 = h2
Y
hh X
(–h,k) (h,k)
X′
hh
(–h,–k) (h,–k)
Y′
(4) If the circle touches both the axes then its equation is (Four cases) Y
(x ± r)2 + (y ± r)2 = r 2 (–r,r) (r,r)
X′ X
(–r,–r) (r,–r)
Y′
(5) If the circle touches x- axis at origin (Two cases) Y(0,k) X
x 2 + (y ± k)2 = k 2 X′ (0,–k)
⇒ x 2 + y 2 ± 2ky = 0
(6) If the circle touches y-axis at origin (Two cases) Y′
(x ± h)2 + y 2 = h2 Y
⇒ x 2 + y 2 ± 2xh = 0 X′ (–h,0) (h,0) X
Y′
(7) If the circle passes through origin and cut intercepts of a and b on axes, the equation of circle is (Four
cases)
x 2 + y 2 − ax − by = 0 and centre is (a / 2, b / 2) Y
b
aX
Note : Circumcircle of a triangle : If we are given sides of a triangle, then first we should find vertices
then we can find the equation of the circle using general form.
Alternate : If equation of the sides are L1 = 0, L2 = 0 and L3 = 0 , then equation of circle is
(L1. L2 ) + λ (L2. L3 ) + µ(L3 .L1) = 0 , where λ and µ are the constant which can be found out by
the conditions, coefficient of x 2 = coefficient of y 2 and coefficient of xy = 0
If the triangle is right angled then its hypotenuse is the diameter of the circle. So using diameter
form we can find the equation.
Circumcircle of a square or a rectangle : Diagonals of the square and rectangle will be
diameters of the circumcircle. Hence finding the vertices of a diagonal, we can easily determine
the required equation.
Alternate : If sides of a quadrilateral are L1 = 0, L2 = 0, L3 = 0 and L4 = 0 . Then equation of
circle is L1L3 + λL2 L4 = 0, where λ is a constant which can be obtained by the condition of circle.
If a circle is passing through origin then constant term is absent i.e. x 2 + y 2 + 2gx + 2 fy = 0
If the circle x 2 + y 2 + 2gx + 2 fy + c = 0 touches X-axis, then − f = g 2 + f 2 − c or g 2 = c
If the circle x 2 + y 2 + 2gx + 2 fy + c = 0 touches Y-axis, then − g = g 2 + f 2 − c or f 2 = c
If the circle x 2 + y 2 + 2gx + 2 fy + c = 0 touches both axes, then − g = − f = g 2 + f 2 − c or g2= f 2= c
Example: 1 A point P moves in such a way that the ratio of its distances from two coplanar points is always fixed number (≠ 1) . Then
its locus is [IIT 1970]
(a) Straight line (b) Circle (c) Parabola (d) A pair of straight lines
Solution: (b) Let two coplanar points are (0, 0) and (a, 0) and coordinates of point P is (x, y).
Under given conditions, we get x2 + y2 = λ (where λ is any number and λ ≠ 1 )
(x − a)2 + y2
⇒ x2 + y2 = λ2 [(x − a)2 + y2] ⇒ x 2 + y2 + λ2 (a2 − 2ax) = 0 , which is equation of a circle.
λ2 −
1
Example : 2 The lines 2x − 3y = 5 and 3x − 4y = 7 are the diameters of a circle of area 154 square units. The equation of the circle is
Solution : (b)
Example : 3 [IIT 1989; AIEEE 2003; DCE 2001]
Solution : (d)
Example : 4 (a) x2 + y2 + 2x − 2y = 62 (b) x2 + y2 − 2x + 2y = 47
Solution : (d)
(c) x2 + y2 + 2x − 2y = 47 (d) x2 + y2 − 2x + 2y = 62
Example : 5
Solution : (a) Centre of circle = Point of intersection of diameters,
On solving equations, 2x − 3y = 5 and 3x − 4y = 7 , we get, (x, y) = (1,−1)
∴ Centre of circle = (1, − 1) . Now area of circle = 154 ⇒ πr2 = 154 ⇒ r = 7
Hence, the equation of required circle is (x − 1)2 + (y + 1)2 = (7)2 ⇒ x2 + y2 − 2x + 2y = 47 .
The equation of a circle with origin as centre passing through the vertices of an equilateral triangle whose median is of
length 3a is [BIT Ranchi 1992; AIEEE 2002]
(a) x2 + y2 = 9a2 (b) x2 + y2 = 16a2 (c) x2 + y2 = a2 (d) None of these
Since the triangle is equilateral, therefore centroid of the triangle is the same as the circumcentre and radius of the circum-
circle = 2 (median) = 2 (3a) = 2a [ Centroid divides median in ratio of 2 : 1]
3 3
Hence, the equation of the circum-circle whose centre is (0, 0) and radius 2a is x 2 + y 2 = (2a)2 ⇒ x 2 + y 2 = 4a 2
A circle of radius 5 units touches both the axes and lies in first quadrant. If the circle makes one complete roll on x-axis
along the positive direction of x-axis, then its equation in the new position is
(a) x 2 + y 2 + 20πx − 10y + 100π 2 = 0 (b) x 2 + y 2 + 20πx + 10y + 100π 2 = 0
(c) x 2 + y 2 − 20πx − 10y + 100π 2 = 0 (d) None of these
The x-coordinate of the new position of the circle is 5 + circumferrence of the first circle = 5 + 10π
The y-coordinate is 5 and the radius is also 5.
Hence, the equation of the circle in the new position is (x − 5 − 10π )2 + (y − 5)2 = (5)2
⇒ x 2 + 25 + 100π 2 − 10x + 100π − 20πx + y 2 + 25 − 10y = 25
⇒ x 2 + y 2 − 20πx − 10x − 10y + 100π 2 + 100π + 25 = 0
The abscissae of A and B are the roots of the equation x2 + 2ax − b2 = 0 and their ordinates are the roots of the equation
y 2 + 2by − q 2 = 0. The equation of the circle with AB as diameter is [IIT 1984]
(a) x 2 + y 2 + 2ax + 2by − b 2 − q 2 = 0 (b) x 2 + y 2 + 2ax + by − b 2 − q 2 = 0
(c) x 2 + y 2 + 2ax + 2by + b 2 + q 2 = 0 (d) None of these
Let x1, x 2 and y1, y2 be roots of x 2 + 2ax − b 2 = 0 and y 2 + 2by − q 2 = 0 respectively.
Then, x1 + x 2 = −2a, x1 x 2 = −b 2 and y1 + y2 = −2b, y1y2 = −q 2
The equation of the circle with A(x1, y1 ) and B(x 2 , y2 ) as the end points of diameter is
(x − x1 ) (x − x 2 ) + (y − y1 ) (y − y2 ) = 0
x 2 + y 2 − x(x1 + x 2 ) − y(y1 + y2 ) + x1 x 2 + y1y2 = 0 ; x 2 + y 2 + 2ax + 2by − b 2 − q 2 = 0
Example : 6 The equation of a circle of radius 1 touching the circles x 2 + y 2 − 2| x|= 0 is
(a) x 2 + y 2 + 2 3x − 2 = 0 (b) x 2 + y 2 − 2 3y + 2 = 0
(c) x 2 + y 2 + 2 3y + 2 = 0 (d) x 2 + y 2 + 2 3x + 2 = 0
Solution : (b,c) The given circles are x 2 + y 2 − 2x = 0, x > 0, and x 2 + y 2 + 2x = 0, x < 0.
From the figure, the centres of the required circles will be (0, 3) and (0, − 3) . 1 1
∴ The equations of the circles are (x − 0)2 + (y 3)2 = 12. 11
⇒ x2 + y2 + 3 2 3y = 1 11 (1,0)
(–1,0)
⇒ x2 + y2 2 3y + 2 = 0
Example : 7 If the line x + 2by + 7 = 0 is a diameter of the circle x 2 + y 2 − 6x + 2y = 0, then b = [MP PET 1991]
Solution : (d)
Example : 8 (a) 3 (b) – 5 (c) – 1 (d) 5
Solution : (b)
Here the centre of circle (3, –1) must lie on the line x + 2by + 7 = 0. Therefore, 3 − 2b + 7 = 0 ⇒ b = 5
Example : 9
The centre of the circle r 2 = 2 − 4r cosθ + 6 r sinθ is
(a) (2, 3) (b) (– 2, 3) (c) (– 2, – 3) (d) (2, – 3)
Let r cos θ = x and r sin θ = y
Squaring and adding, we get r 2 = x 2 + y 2 .
Putting these values in given equation, x 2 + y 2 = 2 − 4 x + 6y ⇒ x2 + y2 + 4 x − 6y − 2 = 0
Hence, centre of the circle = (–2, 3)
The number of integral values of λ for which x 2 + y 2 + λx + (1 − λ)y + 5 = 0 is the equation of a circle whose radius
cannot exceed 5, is
(a) 14 (b) 18 (c) 16 (d) None of these
Solution : (c) Centre of circle = − λ , − (1 − λ) ; Radius of circle = λ 2 + 1 −λ 2 − 5 ≤5
2 2 2 2
⇒ 2λ2 − 2λ − 119 ≤ 0 , ∴ 1− 239 ≤ λ≤ 1+ 239
2 2
⇒ − 7.2 ≤ λ ≤ 8.2 (Nearly). ∴ λ = − 7, − 6, ................,7, 8. Hence number of integral values of λ is 16.
Example : 10 Let f(x, y) = 0 be the equation of a circle. If f(0, λ) = 0 has equal roots λ = 2, 2 and f(λ, 0) = 0 has roots λ = 4 ,5, then
Solution : (b) 5
the centre of the circle is
(a) 2, 29 (b) 29 , 2 (c) − 2, 29 (d) None of these
10 10 10
f(x, y) ≡ x 2 + y 2 + 2gx + 2 fy + c = 0
Now, f(0, λ) ≡ λ2 + 2 fλ + c = 0 and its roots are 2, 2. ∴ 2 + 2 = −2 f, 2 × 2 = c, i.e. f = −2, c = 4
f(λ, 0) ≡ λ2 + 2gλ + c = 0, and its roots are 4 , 5.
5
∴ 4 + 5 = −2g, 4 × 5 = c, i.e., g = −29 , c = 4 . Hence, centre of the circle = (−g, − f) = 29 , 2 .
5 5 10 10
Example : 11 If the lines 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0 are tangents to a circle, then the radius of the circle is
Solution : (b)
[IIT 1984; MP PET 1994, 2002; Rajasthan PET 1995, 97; Kurukshetra CEE 1998]
(a) 3/2 (b) 3/4 (c) 1/10 (d) 1/20
Since both tangents are parallel to each other. The diameter of the circle is perpendicular distance between the parallel
lines (tangents) 3x − 4y + 4 = 0 and 3x − 4y − 7 = 0 and so it is equal to 4 7/2 = 3 .
2 + 9 + 16 2
9 + 16
Hence radius of circle is 3 . (0, 1) 3x – 4y+4=0
4
Alternative method : Perpendicular distance = 3(0) − 4(1) − 7 / 2 = 3 ,
5 2
3x – 4y – 7/2=0
i.e., Diameter = 3
2
Hence radius of circle is 3 .
4
Intercepts on the Axes.
The lengths of intercepts made by the circle x 2 + y2 + 2gx + 2 fy + c = 0 with X and Y axes are 2 g2 − c and
2 f 2 − c respectively.
Let the equation of circle be x 2 + y2 + 2gx + 2 fy + c = 0 ......(i)
Length of intercepts on x-axis and y-axis are |AB|=|x2 − x1| and |CD|=|y2 − y1| respectively.
The circle intersects the x-axis, when y = 0, then x 2 + 2gx + c = 0 . Y
Since the circle intersects the x-axis at A(x1,0) and B(x2,0) .
Then x1 + x 2 = −2g, x1 x 2 = c (0,y2)D
∴ |AB|=|x2 − x1|= (x2 + x1)2 − 4 x1x2 = 2 (g 2 − c)
As the circle intersects the y-axis, when x = 0, then y2 + 2 fy + c = 0 (0,y1)C (xB2,0) X
O (Ax1,0)
Since the circle intersects the y-axis at C (0, y1) and D (0, y2), then y1 + y2 = −2 f, y1y2 = c
∴ |CD|=|y2 − y1|= (y2 + y1 )2 − 4y2y1 = 2 ( f 2 − c) .
Note : If g 2 > c, then the roots of the equation x 2 + 2gx + c = 0 are real and distinct, so the circle
x 2 + y2 + 2gx + 2 fy + c = 0 meets the x-axis in two real and distinct points and the length of the
intercept on x-axis is 2 g 2 − c.
If g 2 = c, then the roots of the equation x 2 + 2gx + c = 0 are real and equal, so the circle touches
x-axis and the intercept on x-axis is zero.
If g 2 < c, then the roots of the equation x 2 + 2gx + c = 0 are imaginary, so the circle
x 2 + y2 + 2gx + 2 fy + c = 0 does not meet x-axis in real points.
Similarly, the circle x 2 + y2 + 2gx + 2 fy + c = 0 cuts the y-axis in real and distinct points, touches
or does not meet in real points according as f 2 >, = or < c .
Position of a point with respect to a Circle.
A point (x1,y1) lies outside, on or inside a circle S ≡ x 2 + y2 + 2gx + 2 fy + c = 0 according as
S1 ≡ x12 + y12 + 2gx1 + 2 fy1 + c is positive, zero or negative i.e.,
S1 > 0⇒ Point is outside the circle.
S1 = 0 ⇒ Point is on the circle.
S1 < 0 ⇒ Point is inside the circle.
(1) The least and greatest distance of a point from a circle : Let S = 0 be a circle and A(x1,y1) be a
point. If the diameter of the circle is passing through the circle at P and Q, then Q
AP = AC − r = least distance ; AQ = AC + r = greatest distance
rC
where 'r' is the radius and C is the centre of the circle.
P
A
Example : 12 The number of points with integral coordinates that are interior to the circle x 2 + y 2 = 16 is
Solution : (c)
(a) 43 (b) 49 (c) 45 (d) 51
Example : 13
Solution : (a) The number of points is equal to the number of integral solutions (x, y) such that x 2 + y 2 < 16 .
So, x, y are integers such that −3 ≤ x ≤ 3, − 3 ≤ y ≤ 3 satisfying the inequation x 2 + y 2 < 16 . The number of selections of
values of x is 7, namely –3, –2, –1, 0, 1, 2, 3. The same is true for y. So the number of ordered pairs (x, y) is 7×7. But
(3, 3), (3, –3), (–3, 3), (–3, –3) are rejected because they do not satisfy the inequation x 2 + y 2 < 16.
So the number of points is 45.
The range of values of a for which the point (a, 4) is outside the circles x 2 + y 2 + 10x = 0 and x 2 + y 2 − 12x + 20 = 0 is
(a) (−∞,− 8) ∪ (−2, 6) ∪ (6, + ∞) (b) (– 8, – 2)
(c) (−∞,− 8) ∪ (−2, + ∞) (d) None of these
For circle, x 2 + y 2 + 10x = 0 ;
a 2 + (4)2 + 10a > 0 ⇒ a 2 + 10a + 16 > 0 ⇒ (a + 8) (a + 2) > 0 ⇒ a < −8 or a > −2 …..(i)
For circle, x 2 + y 2 − 12x + 20 = 0 ; a 2 + (4)2 − 12a + 20 > 0 ⇒ a 2 − 12a + 36 > 0
⇒ (a − 6)2 > 0 ⇒ a ∈ R ∼ {6} …..(ii)
Taking common values from (i) and (ii), a ∈(−∞,− 8) ∪ (−2, 6) ∪ (6, + ∞) .
Intersection of a Line and a Circle. …..(i)
Let the equation of the circle be x 2 + y2 = a2
and the equation of the line be y = mx + c …..(ii)
From (i) and (ii), x 2 + (mx + c)2 = a2 or (1 + m2)x 2 + 2mcx + c 2 − a2 = 0 …..(iii)
Case I: When points of intersection are real and distinct. In this case (iii) has two distinct roots.
∴ B2 − 4 AC > 0 or 4m2c 2 − 4(1 + m2)(c 2 − a 2) > 0 or a2 > c2 P
1 + m2 O
or a > |c | = length of perpendicular from (0, 0) to y = mx + c Q
(1 + m2 )
⇒ a > length of perpendicular from (0, 0) to y = mx + c
Thus, a line intersects a given circle at two distinct points if radius of circle is greater than the length of
perpendicular from centre of the circle to the line.
Case II: When the points of intersection are coincident in this case (iii) has two equal roots.
∴ B2 − 4 AC = 0 ⇒ 4m2c 2 − 4(1 + m2 )(c 2 − a 2 ) = 0
∴ a2 = c2 or a = |c | R
1 + m2 (1 + m2) O
a = length of perpendicular from the point (0, 0) to y = mx + c .
Thus, a line touches the circle if radius of circle is equal to the length of perpendicular from centre of the circle
to the line.
Case III: When the points of intersection are imaginary. In this case (iii) has imaginary roots.
∴ B2 − 4 AC < 0 ⇒ 4m2c 2 − 4(1 + m2 )(c 2 − a 2 ) < 0 , ∴ a2 c2
< 1+ m2
or a < |c | = length of perpendicular from (0, 0) to y = mx + c O
(1 + m2 )
⇒ a < length of perpendicular from (0, 0) to y = mx + c
Thus, a line does not intersect a circle if the radius of circle is less than the length of perpendicular from centre
of the circle to the line.
(1) The length of the intercept cut off from a line by a circle : The length of the intercept cut off from
the line y = mx + c by the circle x 2 + y 2 = a 2 is 2 a 2 (1 + m2 ) − c 2
1 + m2
(2) Condition of tangency : A line L = 0 touches the circle S = 0, if length of perpendicular drawn from the
centre of the circle to the line is equal to radius of the circle i.e. p = r. This is the condition of tangency for the line L = 0.
Circle x 2 + y 2 = a 2 will touch the line y = mx + c if c = ±a 1 + m2
Again, (i) If a 2(1 + m2 ) − c 2 > 0 line will meet the circle at real and different points.
(ii) If c 2 = a 2(1 + m2 ) line will touch the circle.
(iii) If a 2(1 + m2 ) − c 2 < 0 line will meet circle at two imaginary points.
Example : 14 If the straight line y = mx is outside the circle x 2 + y 2 − 20y + 90 = 0, then [Roorkee 1999]
Solution : (d)
(a) m > 3 (b) m < 3 (c) | m | > 3 (d) | m | < 3
Example : 15
Solution : (c) If the straight line y = mx is outside the given circle then perpendicular distance of line from centre of circle > radius of circle
10 > 10 ⇒ (1 + m2 ) < 10 ⇒ m2 < 9 ⇒ |m|< 3
1 + m2
If the chord y = mx + 1 of the circle x2 + y2 = 1 subtends an angle of measure 45o at the major segment of the circle then
value of m is [AIEEE 2002]
(a) 2
(b) – 2 (c) ± 1 (d) None of these
Given circle is x 2 + y 2 = 1, C (0,0) and radius = 1 and chord is y = mx + 1
cos 45o = CP ; CP = Perpendicular distance from (0, 0) to chord y = mx +1 45°
CR
CP = 1 (CR = radius =1) C(0,0)
m2 +1 1
y=mx+1
1 45°45°
m2 + 1 1= 1 ⇒ m2 + 1 = 2 ⇒ m = ±1. PR
1 2 m2 + 1
cos 45o = ⇒
Tangent to a Circle at a given Point.
The limiting position of the line PQ, when Q moves towards P and ultimately coincides with P, is called the
tangent to the circle at the point P. The point P is called the point of contact.
(1) Point form
(i) The equation of tangent at (x1, y1) to circle x 2 + y 2 = a 2 is xx1 + yy1 = a 2 P Q
(ii) The equation of tangent at (x1,y1) to circle x 2 + y 2 + 2gx + 2 fy + c = 0 is Q Q
xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
Tangent Q
Note : For equation of tangent of circle at (x1,y1), substitute xx1 for x 2, yy1 for y2, x + x1 for x, y + y1
2 2
for y and xy1 + x1y for xy and keep the constant as such.
2
This method of tangent at (x1,y1) is applied any conics of second degree. i.e., equation of tangent of
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0 at (x1,y1)
is axx1 + h(xy1 + x1y) + byy1 + g(x + x1) + f(y + y1) + c = 0
(2) Parametric form : Since parametric co-ordinates of circle x 2 + y2 = a2 is (a cosθ ,a sinθ ), then equation
of tangent at (a cosθ ,a sinθ ) is x.a cosθ + y.a sinθ = a2 or x cosθ + y sinθ = a .
(3) Slope form : Let y = mx + c is the tangent of the circle x 2 + y2 = a2 .
∴ Length of perpendicular from centre of circle (0, 0) on line (y = mx + c) = radius of circle
∴ |c| = a ⇒ c = ± a 1 + m2
1 + m2
Substituting this value of c in y = mx + c, we get y = mx ± a 1 + m2 . Which are the required equations of tangents.
Note : The reason why there are two equations y = mx ± a 1 + m2 is that there are two tangents, both
are parallel and at the ends of a diameter.
The line ax + by + c = 0 is a tangent to the circle x 2 + y2 = r 2 if and only if c 2 = r 2(a2 + b2).
The condition that the line lx + my + n = 0 touches the circle x 2 + y2 + 2gx + 2 fy + c = 0 is
(lg + mf − n)2 = (l 2 + m2 )(g 2 + f 2 − c).
Equation of tangent to the circle x 2 + y2 + 2gx + 2 fy + c = 0 in terms of slope is
y = mx + mg − f ± (g 2 + f 2 − c) (1 + m2)
(4) Point of contact : If circle be x 2 + y2 = a2 and tangent in terms of slope be y = mx ± a (1 + m2) ,
Solving x 2 + y2 = a2 and y = mx ± a (1 + m2) simultaneously, we get x = ± am and y = a
(1 + m2) (1 + m2)
Thus, the co-ordinates of the points of contact are ± am , a
(1 + m2)
(1 + m2 )
Alternative method : Let point of contact be (x1,y1) then tangent at (x1,y1) of x 2 + y2 = a2 is
xx1 + yy1 = a 2 . Since xx1 + yy1 = a 2 and y = mx ± a (1 + m2) are identical , ∴ x1 = y1 = + a2
m −1 ± a 1 + m2
∴ x1 = ± am and y1 = a
(1 + m2) (1 + m2)
Thus, the co-ordinates of the points of contact are ± am , a
(1 + m2)
(1 + m2 )
Note : If the line y = mx + c is the tangent to the circle x2 + y2 = r2 then point of contact is given by − mr 2 , r2
c c
If the line ax+by+c = 0 is the tangent to the circle x2+y2=r2 then point of contact is given by − ar 2 , − br 2
c c
Example : 16 The equations to the tangents to the circle x 2 + y 2 − 6x + 4y = 12 which are parallel to the straight line 4x+3y+5=0, are
Solution : (c) [ISM Dhanbad 1973, MP PET 1991]
Example : 17
Solution : (a) (a) 3x − 4y − 19 = 0, 3x − 4y + 31 = 0 (b) 4 x + 3y − 19 = 0, 4 x + 3y + 31 = 0
Example : 18
Solution : (b) (c) 4 x + 3y + 19 = 0, 4 x + 3y − 31 = 0 (d) 3x − 4y + 19 = 0, 3x − 4y + 31 = 0
Let equation of tangent be 4 x + 3y + k = 0, then 9 + 4 + 12 = 4(3) + 3(−2) + k ⇒ 6 + k = ± 25 ⇒ k = 19 and – 31
16 + 9
Hence the equations of tangents are 4 x + 3y + 19 = 0 and 4 x + 3y − 31 = 0
The equations of any tangents to the circle x 2 + y 2 − 2x + 4y − 4 = 0 is
(a) y = m(x − 1) + 3 1 + m2 − 2 (b) y = mx + 3 1 + m2
(c) y = mx + 3 1 + m2 − 2 (d) None of these
Equation of circle is (x − 1)2 + (y + 2)2 = 32 .
As any tangent to x 2 + y 2 = 32 is given by y = mx + 3 1 + m2
Any tangent to the given circle will be y + 2 = m(x − 1) + 3 1 + m2 ⇒ y = m(x − 1) + 3 1 + m2 − 2
If a circle, whose centre is (–1, 1) touches the straight line x + 2y + 12 = 0, then the coordinates of the point of contact are
[MP PET 1998]
(a) − 7 , − 4 (b) − 18 , − 21 (c) (2, –7) (d) (– 2, – 5)
2 5 5
Let point of contact be P(x1, y1 ). ….. (i) P(x1,y1)
This point lies on the given line , ∴ x1 + 2y1 = −12
Gradient of OP = m1 = y1 − 1 , Gradient of x + 2y + 12 = m2 = − 1 O(–1,1)
x1 + 1 2
Both are perpendicular, ∴ m1m2 = −1
⇒ y1 − 1 −1 = − 1 ⇒ y1 − 1 = 2x1 + 2 ⇒ 2x1 − y1 = −3 ….. (ii)
x1 + 1 2
On solving the equation (i) and (ii), (x1, y1) = −18 , −21
5 5
Length of Tangent.
From any point, say P(x1,y1) two tangents can be drawn to a circle which are real, coincident or imaginary
according as P lies outside, on or inside the circle. Q
Le PQ and PR be two tangents drawn from P(x1,y1) to the circle S1
x 2 + y2 + 2gx + 2 fy + c = 0. Then PQ =PR is called the length of tangent drawn
P
(x1,y1)
from point P and is given by PQ = PR = x 2 + y12 + 2gx1 + 2 fy1 +c = S1 R
1
Pair of Tangents.
From a given point P(x1,y1) two tangents PQ and PR can be drawn to the P Q
circle S = x 2 + y2 + 2gx + 2 fy + c = 0. Their combined equation is SS1 = T 2. (x1,y1) R
Where S = 0 is the equation of circle, T = 0 is the equation of tangent at
(x1,y1) and S1 is obtained by replacing x by x1 and y by y1 in S.
Power of Point with respect to a Circle.
Let P(x1,y1) be a point outside the circle and PAB and PCD drawn two secants. The power of P(x1,y1) with
respect to S ≡ x 2 + y2 + 2gx + 2 fy + c = 0 is equal to PA . PB which is TB
A
x12 + y12 + 2gx1 + 2 fy1 + c = 0 ⇒ S1 = 0
∴ Power remains constant for the circle i.e., independent of A and B.
∴ PA. PB = PC.CD = (PT)2 = S1 = ( S1 )2 P(x1,y1) C D
∴ PA . PB = ( S1 )2 = square of the length of tangent.
Note : If P is outside, inside or on the circle then PA . PB is +ve, –ve or zero respectively.
Important Tips
• The length of the tangent drawn from any point on the circle x 2 + y2 + 2gx + 2 fy + c1 = 0 to the circle x 2 + y2 + 2gx + 2 fy + c = 0
is c − c1 .
• If two tangents drawn from the origin to the circle x 2 + y2 + 2gx + 2 fy + c = 0 are perpendicular to each other, then g 2 + f 2 = 2c.
• If the tangent to the circle x 2 + y2 = r 2 at the point (a, b) meets the coordinate axes at the points A and B and O is the origin, then
the area of the triangle OAB is r 4 .
2ab −
• If θ is the angle subtended at P (x1, y1) by the circle S= x2 + y2 + 2gx + 2 fy + c = 0, then cot θ = S1
2 g2 + f2 − c
• The angle between the tangents from (α, β ) to the circle x2 + y2 = a2 is 2 tan−1 α2 a − a2 .
+ β2
• If OA and OB are the tangents from the origin to the circle x 2 + y2 + 2gx + 2 fy + c = 0 and C is the centre of the circle, then the area
of the quadrilateral OACB is c (g 2 + f 2 − c).
Example : 19 If the distances from the origin to the centres of three circles x2 + y2 + 2λix − c2 = 0 (i = 1, 2, 3) are in G.P. then the lengths
Solution : (b) of the tangents drawn to them from any point on the circle x 2 + y 2 = c 2 are in
Example : 20 (a) A.P. (b) G.P. (c) H.P. (d) None of these
Solution : (c)
The centres of the given circles are (−λi, 0) (i = 1, 2, 3)
Example : 21
Solution : (c) The distances from the origin to the centres are λi (i = 1, 2, 3). It is given that λ22 = λ1λ3 .
Let P(h, k) be any point on the circle x 2 + y 2 = c 2 , then, h2 + k 2 = c 2
Now, Li = length of the tangent from (h, k) to x 2 + y2 + 2λi x − c 2 = 0
= h2 + k 2 + 2λih − c 2 = c 2 + 2λih − c 2 = 2λih [ h2 + k2 = c2 and i = 1, 2, 3]
Therefore, L22 = 2λ2h = 2h( λ1λ3 ) [ λ22 = λ1λ3 ]
= 2hλ1 2hλ3 = L1L3 . Hence, L1, L2 , L3 are in G.P.
From a point on the circle x 2 + y 2 = a 2 , two tangents are drawn to the circle x 2 + y 2 = a 2 sin 2 α. The angle between
them is [Rajasthan PET 2002]
(a) α (b) α (c) 2α (d) None of these
2
Let any point on the circle x 2 + y 2 = a 2 be (a cos t, a sin t) and ∠OPQ = θ Q
Now; PQ = length of tangent from P on the circle x2 + y2 = a2 sin2 α O θP
∴ PQ = a2 cos2 t + a2 sin2 t − a2 sin2 α = a cos α
R
OQ = Radius of the circle x 2 + y 2 = a 2 sin 2 α
OQ = a sin α , ∴ tan θ = OQ = tan α ⇒ θ =α; ∴ Angle between tangents ∠QPR = 2α.
PQ
Alternative Method : We know that, angle between the tangent from (α, β ) to the circle x 2 + y 2 = a 2 is
2 tan−1 a . Let point on the circle x2 + y2 = a2 be (a cos t, a sin t)
α2 + β 2 − a2
Angle between tangent = 2 tan−1 a sin α = 2 tan −1 a sin α = 2α
+ a2 sin2 a cos α
a2 cos2 t t − a2 sin2 α
Two tangents to the circle x 2 + y 2 = 4 at the points A and B meet at P (– 4, 0). The area of quadrilateral PAOB, where O
is the origin, is
(a) 4 (b) 6 2 (c) 4 3 (d) None of these
Clearly, sinθ = 2 = 1 , ∴ θ = 30o. So area (∆POA) = 1 . 2 . 4 . sin 60 o
4 2 2
A
∴ Area (quadrilateral PAOB) = 2. 1 . 2. 4 sin 60 o = 8. 3 =4 3. Pθ 2
2 2 (–4,0) 2 O(0,0)
2
B
Trick : Area of quadrilateral = r S1 = 2 12 = 4 3
Example : 22 The angle between a pair of tangents drawn from a point P to the circle x 2 + y 2 + 4 x − 6y + 9 sin 2 α + 13 cos 2 α = 0 is
2α. The equation of the locus of the point P is [IIT 1996]
(a) x 2 + y 2 + 4 x − 6y + 4 = 0 (b) x 2 + y 2 + 4 x − 6y − 9 = 0
(c) x 2 + y 2 + 4 x − 6y − 4 = 0 (d) x 2 + y 2 + 4 x − 6y + 9 = 0
Solution : (d) The centre of the circle x 2 + y 2 + 4 x − 6y + 9 sin 2 α + 13 cos 2 α = 0 is C (−2, 3) and its radius is
22 + (−3)2 − 9 sin 2 α − 13 cos 2 α = 4 + 9 − 9 sin 2 α − 13 cos 2 α = 2 sinα
Let P (h, k) be any point on the locus. The ∠APC = α. Also ∠PAC = π / 2 i.e. triangle APC is a right angle triangle.
Thus sin α = AC = 2 sinα A
PC (h + 2)2 + (k − 3)2
⇒ (h + 2)2 + (k − 3)2 = 2 ⇒ (h + 2)2 + (k − 3)2 = 4 α C
P(h,k) (–2,3)
or h2 + k 2 + 4h − 6k + 9 = 0 B
Thus the required equation of the locus is x 2 + y 2 + 4 x − 6y + 9 = 0 .
Normal to a Circle at a given Point.
The normal of a circle at any point is a straight line, which is perpendicular to the tangent at the point and
always passes through the centre of the circle.
(1) Equation of normal: The equation of normal to the circle x 2 + y2 + 2gx + 2 fy + c = 0 at any point
(x1,y1) is y − y1 = y1 + f (x − x1) or x − x1 = y − y1
x1 + g x1 + g y1 + f
P Tangent
Normal 90°
Note : The equation of normal to the circle x2 + y2 = a2 at any point (x1,y1) is xy1 − x1y = 0 or x = y
x1 y1
• The equation of any normal to the circle x 2 + y2 = a2 is y = mx where m is the slope of normal.
• The equation of any normal to the circle x 2 + y2 + 2gx + 2 fy + c = 0 is y + f = m(x + g). where m
is the slope of normal.
• If the line y = mx + c is a normal to the circle with radius r and centre at (a, b) then b = ma + c .
(2) Parametric form : Since parametric co-ordinates of circle x 2 + y2 = a2 is (a cosθ , a sinθ ) .
∴ Equation of normal at (a cosθ , a sinθ ) is x = y or x = y
a cosθ a sinθ cos θ sin θ
or y = x tanθ or y = mx where m = tan θ , which is slope form of normal.
Example : 23 The line lx + my + n = 0 is a normal to the circle x 2 + y 2 + 2gx + 2 fy + c = 0, if [MP PET 1995]
Solution : (a)
(a) lg+ mf − n = 0 (b) lg+ mf + n = 0 (c) lg− mf − n = 0 (d) lg− mf + n = 0
Since normal always passes through centre of circle, therefore (−g, − f) must lie on lx + my + n = 0 . Hence, lg+ mf − n = 0
Chord of Contact of Tangents.
(1) Chord of contact : The chord joining the points of contact of the two tangents to a conic drawn from a
given point, outside it, is called the chord of contact of tangents.
(2) Equation of chord of contact : The equation of the chord of contact (x′,y′) P
of tangents drawn from a point (x1,y1) to the circle x 2 + y2 = a2 is A Chord of
xx1 + yy1 = a 2. (x1,y1) contact
(x″,y″)Q
Equation of chord of contact at (x1,y1) to the circle x 2 + y2 + 2gx + 2 fy + c = 0 is xx1 + yy1 + g (x + x1)
+ f (y + y1) + c = 0.
It is clear from the above that the equation to the chord of contact coincides with the equation of the tangent,
if point (x1,y1) lies on the circle.
The length of chord of contact = 2 r 2 − p2 ; (p being length of perpendicular from centre to the chord)
Area of ∆APQ is given by a(x12 + y12 − a 2 )3 / 2 .
x12 + y12
(3) Equation of the chord bisected at a given point : The equation of the chord of the circle
S ≡ x 2 + y2 + 2gx + 2 fy + c = 0 bisected at the point (x1,y1) is given by T = S'
i.e. xx1 + yy1 + g (x + x1) + f (y + y1) + c = x12 + y12 + 2gx1 + 2 fy1 + c .
Example : 24 Tangents are drawn from any point on the circle x 2 + y 2 = a 2 to the circle x 2 + y 2 = b 2 . If the chord of contact touches
Solution : (b)
Example : 25 the circle x 2 + y 2 = c 2 , a > b, then [MP PET 1999, Rajasthan PET 1999]
Solution : (b)
(a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) a, c, b are in G.P.
Example : 26
Solution : (c) Chord of contact of any point (a cosθ , a sinθ ) on 1st circle with respect to 2nd circle is ax cosθ + ay sinθ = b 2
This chord touches the circle x 2 + y 2 = c 2 ,
Hence, Radius = Perpendicular distance of chord from centre.
c = b2 ⇒ b2 = ac .Hence a,b,c are in G.P.
a cos2 θ + sin2 θ
The area of the triangle formed by the tangents from the point (4, 3) to the circle x 2 + y 2 = 9 and the line joining their
points of contact is [IIT 1981, MP PET 1991]
(a) 25 sq. units (b) 192 sq. units (c) 384 sq. units (d) None of these
192 25 25
The equation of the chord of contact of tangents drawn from P (4, 3) to x 2 + y 2 = 9 is 4 x + 3y = 9. The equation of OP
is y = 3 x. Q
4
Now, OM = (length of the perpendicular from (0, 0) on 4x + 3y − 9 = 0) = 9 P M (0,0)
5 (4,3) R O
∴ QR = 2. QM = 2 OQ2 − OM 2 =2 9− 81 = 24 x2+y2=9
25 5
Now, PM = OP − OM = 5− 9 = 16 .So, Area of ∆ PQR = 1 24 16 = 192 sq. units
5 5 2 5 5 25
The locus of the middle points of those chords of the circle x 2 + y 2 = 4 which subtend a right angle at the origin is
[MP PET 1990; IIT 1984; Rajasthan PET 1997; DCE 2000, 01]
(a) x 2 + y 2 − 2x − 2y = 0 (b) x 2 + y 2 = 4 (c) x 2 + y 2 = 2 (d) (x − 1)2 + (y − 2)2 = 5
Let the mid-point of chord is (h, k). Also radius of circle is 2. Therefore Y
B
OC = cos 45o ⇒ h2 + k2 = 1 ⇒ h2 + k2 = 2 C (h,k)
OB 2 2 45° AX
O
Hence locus is x 2 + y 2 = 2
Example : 27 If two distinct chords, drawn from the point (p, q) on the circle x 2 + y 2 = px + qy (where p, q ≠ 0) are bisected by the
Solution : (d)
x-axis, then [IIT 1999]
(a) p 2 = q 2 (b) p 2 = 8q 2 (c) p 2 < 8q 2 (d) p 2 > 8q 2
Let (h, 0) be a point on x-axis, then the equation of chord whose mid-point is (h, 0) will be
xh − 1 p (x + h) − 1 q (y + 0) = h2 − ph . This passes through (p, q), hence ph − 1 p(p + h) − 1 q.q = h2 − ph
2 2 2 2
⇒ ph − 1 p2 − 1 ph − 1 q 2 = h2 − ph ⇒ h2 − 3 ph + 1 (p 2 + q2) = 0 ; h is real, hence B2 − 4 AC > 0
2 2 2 2 2
∴ 9 p2 − 4. 1 (p 2 + q2) > 0 ⇒ 9p2 − 8(p2 + q2) > 0 ⇒ p2 − 8q2 > 0 ⇒ p2 > 8q2
4 2
Director Circle.
The locus of the point of intersection of two perpendicular tangents to a circle is called the Director circle.
Let the circle be x 2 + y2 = a2 , then equation of the pair of tangents to a 90° P(x1,y1)
circle from a point is (x 2 + y 2 − a 2 ) (x12 + y12 − a 2 ) = (xx1 + yy1 − a 2 )2 . If this
represents a pair of perpendicular lines, coefficient of x 2 + coefficient of y2 = 0
i.e. (x12 + y12 − a 2 − x12 ) + (x12 + y12 − a 2 − y12 ) = 0 ⇒ x12 + y12 = 2a 2
Hence the equation of director circle is x 2 + y2 = 2a2 .
Obviously director circle is a concentric circle whose radius is 2 times the radius of the given circle.
Note : Director circle of circle x 2 + y2 + 2gx + 2 fy + c = 0 is x 2 + y2 + 2gx + 2 fy + 2c − g 2 − f 2 = 0 .
Diameter of a Circle.
The locus of the middle points of a system of parallel chords of a circle is called a diameter of the circle.
The equation of the diameter bisecting parallel chords y = mx + c (c is a Diameter
B x+my=0
parameter) of the circle x 2 + y2 = a2 is x + my = 0.
O
Note : The diameter corresponding to a system of parallel chords of a P(h,k)
y=mx+c
circle always passes through the centre of the circle and is
perpendicular to the parallel chords. A
Example : 28 A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the equation 2x − y = 2.
Solution : (b) Then the equation of the circle is
(a) x 2 + y 2 + 2x − 1 = 0 (b) x 2 + y 2 − 2x − 1 = 0 (c) x 2 + y 2 − 2y − 1 = 0 (d) None of these.
Example : 29
Solution : (c) The line joining (4, 3) and (2, 1) is also along a diameter. So, the centre is the intersection of the diameters 2x − y = 2
and y − 3 = (x − 4) . Solving these, the centre = (1, 0)
∴ Radius = Distance between (1, 0) and (2, 1) = 2.
∴ Equation of circle (x − 1)2 + y 2 = ( 2)2 ⇒ x 2 + y 2 − 2x − 1 = 0
The diameter of the circle x 2 + y 2 − 4 x + 2y − 11 = 0 corresponding to a system of chords parallel to the line
x − 2y + 1 = 0
(a) x − 2y + 3 = 0 (b) 2x − y + 3 = 0 (c) 2x + y − 3 = 0 (d) None of these
The centre of the given circle is (2, –1) the equation of the line perpendicular to chord x − 2y + 1 = 0 is 2x + y + k = 0
Since the line passes through the point (2, –1) therefore k = – 3. The equation of diameter is 2x + y − 3 = 0.
Pole and Polar.
Let P (x1,y1) be any point inside or outside the circle. Draw chords AB and A' B' passing through P. If
tangents to the circle at A and B meet at Q (h, k), then locus of Q is called the polar of P with respect to circle and P
is called the pole and if tangents to the circle at A' and B' meet at Q', then the straight line QQ' is polar with P as its pole.
If circle be x 2 + y2 = a2 then AB is the chord of contact of Q (h, k), hx + ky = a2 is its equation. But
P (x1,y1) lies on AB, ∴ hx1 + ky1 = a 2 .
Q′ Q(h,k) A
B′ Q′ A′
A′ Polar Polar PP(oxle1,y1)
Pole
B B
P(x1,y1) A B′
Q(h,k)
Hence, locus of Q (h, k) is xx1 + yy1 = a2 , which is polar of P (x1,y1) with respect to the circle x 2 + y2 = a2 .
(1) Coordinates of pole of a line : The pole of the line lx + my + n = 0 with respect to the circle
x 2 + y2 = a2 . Let pole be (x1,y1), then equation of polar with respect to the circle x 2 + y2 = a2 is
xx1 + yy1 − a2 = 0 , which is same as lx + my + n = 0
Then x1 = y1 = a2 , ∴ x1 a 2l and y1 = − a 2m . Hence, the required pole is − a 2l ,− a 2m .
l m −n =− n n n n
(2) Properties of pole and polar
(i) If the polar of P (x1,y1) w.r.t. a circle passes through Q(x2,y2) then the polar of Q will pass through P and
such points are said to be conjugate points.
(ii) If the pole of the line ax + by + c = 0 w.r.t. a circle lies on another line a1x + b1y + c1 = 0; then the pole of
the second line will lie on the first and such lines are said to be conjugate lines.
(iii) The distance of any two points P (x1,y1) and Q(x2,y2) from the centre of a circle is proportional to the
distance of each from the polar of the other.
(iv) If O be the centre of a circle and P any point, then OP is perpendicular to the polar of P.
(v) If O be the centre of a circle and P any point, then if OP (produced, if necessary) meet the polar of P in Q,
then OP . OQ = (radius)2.
Note : Equation of polar is like as equation of tangent i.e., T = 0 (but point different)
Equation of polar of the circle x 2 + y2 + 2gx + 2 fy + c = 0 with respect to (x1,y1) is
xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
If the point P is outside the circle then equation of polar and chord of contact will coincide. In this
case the polar cuts the circle at two points.
If the point P is on the circle then equation of polar, chord of contact and tangent at P will coincide.
So in this case the polar touches the circle.
If the point P is inside the circle (not its centre) then only its polar will exist. In this case the polar is
outside the circle. The polar of the centre lies at infinity.
If a triangle is like that its each vertex is a pole of opposite side with respect to a circle then it is
called self conjugate triangle.
Example : 30 The polar of the point 5, − 1 with respect to circle (x − 2)2 + y2 =4 is [Rajasthan PET 1996]
Solution : (b) 2
Example : 31 (d) x − 10y − 2 = 0
Solution : (c) (a) 5x − 10y + 2 = 0 (b) 6x − y − 20 = 0 (c) 10x − y − 10 = 0
The polar of the point 5, − 1 is xx1 + yy1 + g(x + x1) + f(y + y1 ) + c =0
2
⇒ 5x − 1 y − 2(x + 5) + 0 + 0 = 0 ⇒ 3x − y − 10 = 0 ⇒ 6x − y − 20 = 0 .
2 2
The pole of the straight line 9x + y − 28 = 0 with respect to circle 2x 2 + 2y 2 − 3x + 5y − 7 = 0 is
[Rajasthan PET 1990, 99; MNR 1984; UPSEAT 2000]
(a) (3, 1) (b) (1, 3) (c) (3, – 1) (d) (– 3, 1)
Equation of given circle is x2 + y2 − 3 x + 5 y − 7 =0
2 2 2
x − 3 2 + y + 5 2 9 25 7 x 3 2 y 5 2 45
4 4 − 16 16 2 4 4 8
⇒ − − = 0 ⇒ − + + − = 0
Put X = x − 3 and Y = y + 5 , we get the equation of circle X2 + Y 2 − 45 =0 and the line 9X + Y − 45 =0
4 4 8 2
Hence pole 9 45 1 × 45 9 1 . But, 9 3 and 1 5 = −1 , hence the pole is (3, – 1).
×8 8 4 4 4 4 4 4
= 45 , = , x= + =3 y= −
45
2 2
Two Circles touching each other.
(1) When two circles touch each other externally : Then distance between their centres = Sum of their
radii i.e., |C1C2|= r1 + r2
In such cases, the point of contact P divides the line joining C1 and C2
internally in the ratio r1 : r2 ⇒ C1 P = r1 r1 r2
C2 P r2 C1 P C2
If C1 ≡ (x1, y1) and C2 ≡ (x2, y2) , then co-ordinate of P is
r1 x2 + r2 x1 , r1y2 + r2y1
r1 + r2 r1 + r2
(2) When two circles touch each other internally : Then distance between their centres = Difference of
their radii i.e., |C1C2|= r1 − r2
In such cases, the point of contact P divides the line joining C1 and C2
externally in the ratio r1 : r2 ⇒ C1 P = r1 r2
C2 P r2
C1 C2 P
r1
If C1 ≡ (x1, y1) and C2 ≡ (x2, y2) , then co-ordinate of P is
r1 x2 − r2 x1 , r1y2 − r2y1
r1 − r2 r1 − r2
Common Tangents to Two circles.
Different cases of intersection of two circles :
Let the two circles be (x − x1)2 + (y − y1)2 = r12 …..(i)
and (x − x2 )2 + (y − y2 )2 = r22 …..(ii)
with centres C1 (x1,y1) and C2 (x2, y2) and radii r1 and r2 respectively. Then following cases may arise :
Case I : When |C1C2|> r1 + r2 i.e., the distance between the centres is greater than the sum of radii.
In this case four common tangents can be drawn to the two circles, in which Direct common
two are direct common tangents and the other two are transverse common tangents
tangents.
r1 r2
C1 T C2 D
Transvers common
tangents
Case II : When |C1C2 |= r1 + r2 i.e., the distance between the centres is equal to the sum of radii.
In this case two direct common tangents are real and distinct while the Direct common
transverse tangents are coincident. tangents
C1 T C2 D
Transverse common
tangents
Case III : When |C1C2|< r1 + r2 i.e., the distance between the centres is less than sum of radii.
In this case two direct common tangents are real and distinct while the Direct common
transverse tangents are imaginary. tangents
C1 C2 D
Case IV : When |C1C2|= |r1 − r2|, i.e., the distance between the centres is equal to the difference of the radii.
In this case two tangents are real and coincident while the other two Tangent at
tangents are imaginary. the point of
r2 contact
C1 C2 P
r1
Case V : When |C1C2|< |r1 − r2| , i.e., the distance between the centres is less than the difference of the radii.
In this case, all the four common tangents are imaginary.
r2
C1 C2
r1
Note : Points of intersection of common tangents : The points T1 and T2 (points of intersection of
indirect and direct common tangents) divides C1C2 internally and externally in the ratio r1 : r2
respectively.
• Equation of the common tangents at point of contact is S1 − S2 = 0.
If the circle x 2 + y 2 + 2gx + c 2 = 0 and x 2 + y 2 + 2 fy + c 2 = 0 touch each other, then
1 + 1 1 .
g2 f2 = c2
Circle and system of circles
Condition Position Diagram No. of common
tangents
(i) C1C2 > r1 + r2 Do not intersect or one
outside the other 4
C1 T1 C2 T2
(ii) C1C2 < | r1 − r2| One inside the other C2
C1
0
(iii) C1C2 = r1 + r2 External touch 3
1
T1 T2 2
C1 C2
(iv) C1C2 = | r1 − r2| Internal touch
C2
C1
(v) | r1 − r2| < C1C2 < r1 + r2 Intersection at two real
points
C2 T2
C1
Example : 32 If circles x 2 + y 2 + 2ax + c = 0 and x 2 + y 2 + 2by + c = 0 touch each other, then [MNR 1987]
(a) 11 1 (b) 11 1 (c) 1 + 1 = c2 (d) 1 11
a+b = c a2 + b2 = c2 a b a2 + b2 = c
Solution : (d) C1 = (−a, 0), r1 = a 2 − c ; C2 = (0, − b), r2 = b 2 − c ; C1C2 = a 2 + b 2
Circles touch each other, therefore r1 + r2 = C1C2
⇒ a2 − c + b2 − c = a2 + b2 ⇒ a2b2 − b2c − a2c = 0
Multiplying by 1 2 , we get 1 1 = 1 .
a 2b 2c a2 + b2 c
Example : 33 If two circles (x − 1)2 + (y − 3)2 = r 2 and x 2 + y 2 − 8x + 2y + 8 = 0 intersect in two distinct points, then
Solution : (a)
[IIT 1989; Karnataka CET 2002; DCE 2000, 01; AIEEE 2003]
(a) 2 < r < 8 (b) r = 2 (c) r < 2 (d) r > 2
When two circles intersect each other, then difference between their radii < Distance between their centres
⇒ r−3<5 ⇒ r <8 …..(i)
Sum of their radii > Distance between their centres …..(ii)
⇒ r+3>5 ⇒ r >2
Hence by (i) and (ii), 2 < r < 8.
Example : 34 The equation of the circle having the lines x 2 + 2xy + 3x + 6y = 0 as its normals and having size just sufficient to contain
Solution : (b)
the circle x (x − 4) + y (y − 3) = 0 is [Roorkee 1990]
Example : 35
Solution : (d) (a) x 2 + y 2 + 3x − 6y − 40 = 0 (b) x 2 + y 2 + 6x − 3y − 45 = 0
Example : 36
Solution : (b) (c) x 2 + y 2 + 8x + 4y − 20 = 0 (d) x 2 + y 2 + 4 x + 8y + 20 = 0
Example : 37
Solution : (c) Given pair of normals is x 2 + 2xy + 3x + 6y = 0 or (x + 2y)(x + 3) = 0
∴ Normals are x + 2y = 0 and x + 3 = 0
The point of intersection of normals x + 2y = 0 and x + 3 = 0 is the centre of C1
C2
required circle, we get centre C1 = (−3, 3 / 2) and other circle is
x (x − 4) + y (y − 3) = 0 or x 2 + y 2 − 4 x − 3y = 0 …..(i)
Its centre C2 = (2, 3 / 2) and radius r = 4 + 9 = 5
4 2
Since the required circle just contains the given circle (i), the given circle should touch the required circle internally from
inside. Therefore, radius of the required circle =|C1 − C2|+ r = (−3 − 2)2 + 3 − 3 2 + 5 =5 + 5 = 15
2 2 2 2 2
3)2 + y 3 2 15 2 or x 2 + y 2 + 6x − 3y − 45 = 0 .
2 2
Hence, equation of required circle is (x + − =
The equation of the circle which touches the circle x 2 + y 2 − 6x + 6y + 17 = 0 externally and to which the lines
x 2 − 3xy − 3x + 9y = 0 are normals, is [Roorkee 1994]
(a) x 2 + y 2 − 6x − 2y − 1 = 0 (b) x 2 + y 2 + 6x + 2y + 1 = 0
(c) x 2 + y 2 − 6x − 6y + 1 = 0 (d) x 2 + y 2 − 6x − 2y + 1 = 0
Joint equations of normals are x 2 − 3xy − 3x + 9y = 0 ⇒ x (x − 3y) − 3(x − 3y) = 0 ⇒ (x − 3) (x − 3y) = 0
∴ Given normals are x − 3 = 0 and x − 3y = 0 , which intersect at centre of circle whose coordinates are (3, 1).
The given circle is C1 = (3, − 3), r1 = 1 ; C2 = (3, 1), r2 = ?
If the two circles touch externally, then C1C2 = r1 + r2 ⇒ 4 = 1 + r2 ⇒ r2 = 3
∴ Equation of required circle is (x − 3)2 + (y − 1)2 = (3)2 ⇒ x 2 + y 2 − 6x − 2y + 1 = 0
The number of common tangents to the circles x 2 + y 2 = 4 and x 2 + y 2 − 6x − 8y = 24 is [IIT 1998]
(a) 0 (b) 1 (c) 3 (d) 4
Circles S1 ≡ x 2 + y 2 = (2)2 and S2 ≡ (x − 3)2 + (y − 4)2 = (7)2
∴ Centres C1 = (0, 0), C2 = (3, 4) and radii r1 = 2, r2 = 7
C1C2 = (3)2 + (4)2 = 5 , r2 − r1 = 7 − 2 = 5
∴ C1C2 = r2 − r1 i.e. circles touch internally. Hence there is only one common tangent.
There are two circles whose equations are x 2 + y 2 = 9 and x2 + y2 − 8x − 6y + n2 = 0, n∈ Z. If the two circles have
exactly two common tangents, then the number of possible values of n is
(a) 2 (b) 8 (c) 9 (d) None of these
For x 2 + y 2 = 9 , the centre = (0, 0) and the radius = 3
For x 2 + y 2 − 8x − 6y + n2 = 0 . The centre = (4, 3) and the radius = (4)2 + (3)2 − n2
∴ 4 2 + 3 2 − n2 > 0 or n2 < 5 2 or −5 < n < 5.
Circles should cut to have exactly two common tangents.
So, r1 + r2 > C1C2 , ∴ 3 + 25 − n2 > (4)2 + (3)2 or 25 − n2 > 2 or 25 − n2 > 4
∴ n2 < 21 or − 21 < n < 21
Therefore, common values of n should satisfy − 21 < n < 21.
But n∈ Z , So, n = −4, − 3,........3, 4 . ∴ Number of possible values of n = 9.
Common chord of two Circles.
(1) Definition : The chord joining the points of intersection of two given circles is called their common chord.
(2) Equation of common chord : The equation of the common chord of two circles
S1 ≡ x 2 + y 2 + 2g1 x + 2 f1y + c1 = 0 ….(i) PM
and S2 ≡ x 2 + y 2 + 2g 2 x + 2 f2y + c 2 = 0 ….(ii)
is 2x (g1 − g 2 ) + 2y ( f1 − f2 ) + c1 − c 2 = 0 i.e. S1 − S2 = 0. C1 C2
S1=0 S2=0
Q
(3) Length of the common chord : PQ = 2(PM) = 2 C1P 2 − C1M 2
Where C1P = radius of the circle S = 0 and C1M = length of the perpendicular from the centre C1 to the
common chord PQ.
Note : The length of the common chord is 2 r12 − p12 = 2 r22 − p 2 where p1 and p2 are the lengths of
2
perpendicular drawn from the centre to the chord.
While using the above equation of common chord the coefficient of x 2 and y2 in both equation
should be equal.
Two circle touches each other if the length of their common chord is zero.
Maximum length of the common chord = diameter of the smaller circle.
Example : 38 If the common chord of the circles x 2 + (y − λ)2 = 16 and x 2 + y 2 = 16 subtend a right angle at the origin, then λ is
Solution : (c) equal to
Example : 39 (a) 4 (b) 4 2 (c) ± 4 2 (d) 8
Solution : (d)
Example : 40 The common chord of given circles is S1 − S2 = 0
Solution : (c)
⇒ x2 + (y − λ)2 − 16 − {x 2 + y2 − 16} = 0 i.e., y= λ ( λ ≠ 0)
2
The pair of straight lines joining the origin to the points of intersection of y= λ and x2 + y2 = 16 is x2 + y2 = 16 2y 2
2 λ
⇒ λ2 x 2 + (λ2 − 64) y 2 = 0 . These lines are at right angles if λ2 + λ2 − 64 = 0 , i.e., λ = ± 4 2.
Which of the following is a point on the common chord of the circles x 2 + y 2 + 2x − 3y + 6 = 0 and
x 2 + y 2 + x − 8y − 13 = 0 [Karnataka CET 2003]
(a) (1, –2) (b) (1, 4) (c) (1, 2) (d) (1, – 4)
Given circles are, S1 ≡ x 2 + y 2 + 2x − 3y + 6 = 0 ….. (i) and S2 ≡ x 2 + y 2 + x − 8y − 13 = 0 ….. (ii)
∴ Equation of common chord is S1 − S2 = 0
⇒ x + 5y + 19 = 0 , and out of the four given points only point (1, – 4) satisfies it.
If the circle x 2 + y 2 = 4 bisects the circumference of the circle x 2 + y 2 − 2x + 6y + a = 0, then a equals
[Rajasthan PET 1999]
(a) 4 (b) – 4 (c) 16 (d) – 16
The common chord of given circles is S1 − S2 = 0 ⇒ 2x − 6y − 4 − a = 0 …..(i)
Since, x 2 + y 2 = 4 bisects the circumferences of the circle x 2 + y 2 − 2x + 6y + a = 0, therefore (i) passes through the
centre of second circle i.e. (1, – 3). ∴ 2 + 18 – 4 – a = 0 ⇒ a = 16.
Angle of Intersection of Two Circles.
The angle of intersection between two circles S = 0 and S' = 0 is defined as the angle between their tangents
at their point of intersection. A′ B′
If S ≡ x 2 + y 2 + 2g1x + 2 f1y + c1 = 0
S' ≡ x 2 + y 2 + 2g2 x + 2 f2y + c2 = 0 S=0 S′=0
π–rθ1 θ P r2
C1 C2
BA
Q
are two circles with radii r1, r2 and d be the distance between their centres then the angle of intersection θ
between them is given by cosθ = r12 + r12 − d 2 or cosθ = 2(g1g 2 + f1 f2 ) − (c1 + c2 )
2r1r2 2
g12 + f12 − c1 g 2 + f22 − c2
2
(1) Condition of Orthogonality : If the angle of intersection of the two circles is a right angle (θ = 90o ),
then such circles are called orthogonal circles and condition for their orthogonality P
is 2g1g2 + 2 f1 f2 = c1 + c2
Note : When the two circles intersect orthogonally then the length of 90°
C1 C2
tangent on one circle from the centre of other circle is equal to (–g1,–f1) (–g2,–f2)
the radius of the other circle.
Equation of a circle intersecting the three circles x 2 + y2 + 2gi x + 2 fiy + ci = 0 (i = 1, 2,3)
x2 + y2 x y 1
orthogonally is − c1 g1 f1 −1 =0
− c2 g2 f2 −1
− c3 g3 f3 − 1
Example : 41 A circle passes through the origin and has its centre on y = x. If it cuts x 2 + y 2 − 4 x − 6y + 10 = 0 orthogonally, then the
Solution : (c)
equation of the circle is [EAMCET 1994]
Example : 42
Solution : (a) (a) x 2 + y 2 − x − y = 0 (b) x 2 + y 2 − 6x − 4y = 0 (c) x 2 + y 2 − 2x − 2y = 0 (d) x 2 + y 2 + 2x + 2y = 0
Let the required circle be x 2 + y 2 + 2gx + 2 fy + c = 0 .......(i)
This passes through (0, 0), therefore c = 0
The centre (−g, − f) of (i) lies on y = x, hence g = f.
Since (i) cuts the circle x 2 + y 2 − 4 x − 6y + 10 = 0 orthogonally, therefore 2(−2g − 3 f) = c + 10
⇒ −10g =10 ⇒ g = f = −1 ( g = f and c = 0) . Hence the required circle is x 2 + y 2 − 2x − 2y = 0 .
The centre of the circle, which cuts orthogonally each of the three circles x 2 + y 2 + 2x + 17y + 4 = 0,
x 2 + y 2 + 7x + 6y + 11 = 0 and x 2 + y 2 − x + 22y + 3 = 0 is [MP PET 2003]
(a) (3, 2) (b) (1, 2) (c) (2, 3) (d) (0, 2)
Let the circle is x 2 + y 2 + 2gx + 2 fy + c = 0 …..(i)
Circle (i) cuts orthogonally each of the given three circles. Then according to condition 2g1g2 + 2 f1 f2 = c1 + c2
2g + 17 f = c + 4 …..(ii)
7g + 6 f = c + 11 …..(iii)
−g + 22 f = c + 3 …..(iv)
On solving (ii), (iii) and (iv), g = −3, f = −2 . Therefore, the centre of the circle (−g, − f) = (3, 2)
Example : 43 The locus of the centre of a circle which cuts orthogonally the circle x 2 + y 2 − 20x + 4 = 0 and which touches x = 2 is
Solution : (d)
[UPSEAT 2001]
(a) y 2 = 16x + 4 (b) x 2 = 16y (c) x 2 = 16y + 4 (d) y 2 = 16x
Let the circle be x 2 + y 2 + 2gx + 2 fy + c = 0 …..(i)
It cuts the circle x 2 + y 2 − 20x + 4 = 0 orthogonally
∴ 2(−10g + 0 × f) = c + 4 ⇒ − 20g = c + 4 …..(ii)
Circle (i) touches the line x = 2; ∴ x + 0y − 2 = 0
∴ − g + 0 − 2 = g 2 + f 2 − c ⇒ (g + 2)2 = g 2 + f 2 − c ⇒ 4g + 4 = f 2 − c …..(iii)
1
Eliminating c from (ii) and (iii), we get − 16g + 4 = f 2 + 4 ⇒ f 2 + 16g = 0 .
Hence the locus of ( −g, − f) is y 2 − 16x = 0 ⇒ y 2 = 16x.
Family of Circles.
(1) The equation of the family of circles passing through the point of intersection of two given circles S = 0
and S' = 0 is given as
S + λS' = 0 (where λ is a parameter, λ ≠ −1)
S=0 S′=0
S+λS′=0
(2) The equation of the family of circles passing through the point of intersection of circle S = 0 and a line L
= 0 is given as
S + λL = 0 (where λ is a parameter)
S=0 L=0 S+λL=0
(3) The equation of the family of circles touching the circle S = 0 and the line L = 0 at their point of contact P is
S + λL = 0 (where λ is a parameter)
S=0 L=0 S+λL=0
(4) The equation of a family of circles passing through two given points P (x1,y1) and Q(x2,y2) can be
written in the form
x y1 P(x1,y1)
Q(x2,y2
⇒ (x − x1)(x − x2) + (y − y1)(y − y2) + λ x1 y1 1 = 0
x2 y2 1
(where λ is a parameter)
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Free Flip-Book Mathematics class 12th by Study Innovations. 494 Pages
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