Mathematics

Circle Sessions - 3

Session

Session Objectives

Session Objective

1. Equation Of Family Of Circles

2. Equation Of Chord Whose mid –Point is Given

3. Equation Of Chord Of Contact

4. Angle Between Two Circles

5. Orthogonal Intersection

6. Equation Of Common Chord

Equation Of Family Of Circles

Family of circles passing through the intersection of a circle S = 0 and a line L = 0 is

S = 0 L = 0

A

B

S+L = 0

Questions

Illustrative Problem

Write the family of circles passing through the intersection of x2 + y2 –9 = 0 and x + y –1 = 0. Find that member of this family which passes through the origin.

Solution:

Family of required circle is

S + L = 0

Solution Cond.

2 2(x y 9) (x y 1) 0 ...(1)

Since the required circle passes through the origin, we have

(0+0-9) + (0+0-1) = 0

= -9

Substituting value of in (1) we get

2 2x y 9x 9y 0

Equation Of Family Of Circles Passing through points (x1,y1) and (x2,y2)

A(x1,y1)

B(x2,y2)

S (x-x1)(x-x2) + (y-y1)(y-y2) = 0

2 11

2 1

y yL y y 0

x x

Family of circles is S + L = 0

Questions

Illustrative Problem

Find the equation of the circle passing through the (4,1), (6,5) and having its centre on 4x+3y-16=0

Solution:

Circle whose end points of diameter are (4,1), (6,5) is

(x 4)(x 6) (y 1)(y 5) 0

Equation of line passing through (4,1) and (6,5) is 2x-y-7=0.

Therefore family of circles is

2 2x y 10x 6y 29 (2x y 7) 0

Solution Cond.

6centre is 5, lieson 4x 3y 16 0

2

6 264( 5) 3 16 0

2 5

2 2Circle is 5(x y 10x 6y 29) 26(2x y 7) 0

2 25x 5y 2x 56y 37 0

Equation of family of circle which touches a given circle S at a given point (x1,y1)

S=0

L=0

A(x1,y1)

S + L = 0 where L = 0 is tangent at the given point on it.

Questions

Illustrative problem

Find the equation of the circle which touches the circle x2+y2=25 at (3,4) and passes through (1,1).

Solution:

Tangent at (3,4) is 3x+4y-25=0

Therefore family of circle touching x2+y2-25 at (3,4) is

2 2x y 25 (3x 4y 25) 0

Solution Cond.

It passes thorough (1,1)

1 + 1 – 25 + (3x+4y-25) = 0

2318

Therefore the required equation of the circle is

2 2 23x y 25 (3x 4y 25) 0

18

Equation of circles which touches a given line at a given point on it

A(x1,y1)

L = 0

2 21 1S (x x ) (y y ) 0

family of circle is S + L=0

Questions

Illustrative ProblemFind the equation of the circle passing through origin(0,0) and touching the line 2x+y-1=0 at (1,-1)

Solution:

Family of the circles touching 2x+y-1 = 0 at (1,-1) is

2 2(x 1) (y 1) (2x y 1) 0

It passes through (0,0)

= 2

2 2the equation of the circle is x y 2x 4y 0

Equation of Circle Passing through the points of intersection of two circles

S1 + S2 = 0 ( -1)

S1 = 0 S2 = 0

Question

Illustrative Problem

Find the equation of circle passing through origin and the points of intersection of the two circles x2 + y2

- 4x - 6y – 3 = 0 and x2 + y2 + 4x - 2y – 4 = 0

Solution:

Equation of family of circle is

x2 + y2 - 4x - 6y – 3 + (x2 + y2 + 4x - 2y - 4) = 0

It passes through (0,0) 34

=>x2 + y2 - 28x - 18y = 0

Equation of chord whose mid-point is (x1,y1)

S=0

C=(0,0)

A B(x1,y1)D

xx1 + yy1 = x12 + y1

2

T = S1

Questions

Illustrative Problem

Find the equation of the chord of the circle (x-1)2 + (y-2)2 = 4 whose mid – point is (2,1).

Equation of circle is 2 2x y 2x 4y 1 0

x 2 y 1T 2(x) 1(y) 2 4 1 0

2 2

T x y 3 0

Solution: Method 1

2 21S (2) (1) 2(2) 4(1) 1

1T S x y 3 2 x y 1

Illustrative ProblemFind the equation of the chord of the circle (x-1)2 + (y-2)2 = 4 whose mid – point is (2,1).

Solution Method 2:

C=(1,2)

A B(2,1)D

2 1Slope of CD 1

1 2

Slope of AB = 1

Equation of AB is

x – y – 1 = 0

Equation of chord of contact of tangents drawn from a point (x1,y1)

P(x1,y1)

A(x2,y2)

B(x3,y3)

Equation of chord of contact is

xx1+yy1=a2

or

T = 0

Question

Illustrative problem

Find the equation of chord of contact of a point (-2,-3) with respect to circle x2 + y2 - 2x - 6y + 1=0

Required circle is T = 0

x(-2) + y(-3) –2(-2) –6(-3) + 1 = 0

2x + 3y - 23 = 0

Solution :

Angle at which two circles intersect

2 2 21 2

1 2

d r rcos

2r r

S1 =0 S2 = 0

C1 C2

P

r1 r2

d = distance(c1,c2)

Question

Illustrative Problem

Find the angle at which the circles x2 + y2 –8x – 2y - 9 =0 and x2 + y2 + 2x + 8y - 7 = 0 intersect.

C1 = (4,1); r1 = 16 1 9 26

C2 = (-1,-4); r2 = 1 16 7 24

2d 25 25 50

50 26 24cos 0

2 26 24

90

Solution :

Orthogonal Intersection

C1C2

r1 r2

S1=0 S2=0

90

d = distance (c1,c2)

Method 1:

d2 = r12+r2

2

Orthogonal Intersection

C1C2

r1 r2

S1=0 S2=0

90

Method 2:

2g1g2 + 2f1f2 = c1 + c2

Question

Illustrative Problem

If two circles of equal radii ‘a’ with centre (2,3) and (5,6) respectively cut each other orthogonally then find the value of a.

Two circles cut orthogonally

2 2 21 2d r r

2 2d (2 5) (3 6) 3 2

Therefore a2 + a2 = 18 => a = 3

Solution :

Common Chord Of two circles S1 = 0 and S2 = 0

S1 =0 S2 =0

A

B

Equation of common chord is S1 - S2 = 0

Question

Illustrative Problem

Find the equation of common chord of two circles x2 + y2 =25 and 4x2 + 4y2 - 40x + 91=0

2 21here S x y 25 0

2 22

91S x y 10x 0

4

1 2S S 0 40x 191 0

Solution :

Class Test

Class Exercise - 1

A variable chord is drawn through theorigin to the circle x2 + y2 – 2ax = 0.The locus of the centre of the circledrawn on this chord as diameter is

2 2(a) x y ax 0 2 2(b) x y ay 0

2 2(c) x y ax 0 2 2(d) x y ay 0

Solution

Let (h, k) be the centre of the circle.Then (h, k) being the mid-point ofthe chord of the given circle

2 2hx ky a x h h k 2ah.

Since it passes through (0, 0)

2 2ah h k 2ah 2 2h k ah 0

Locus is 2 2x y ax 0

Hence, answer is (c).

Class Exercise - 2

If the circle passes through the point(a, b) and cuts the circle orthogonally,equation of the locus of its centre is

(a) 2ax + 2by = a2 + b2 + k2

2 2 2(b) ax by a b k

2 2 2(c) x y 2ax 2by k 0

2 2 2 2 2(d) x y 2ax 2by a b k 0

Solution

Let (h, k) be the centre

2 22 2 2h k h a k b k

2 2 2 2 2 2 2h k h 2ah a k 2bk b k

2 2 22ah 2bk a b k 0

Locus is 2ax + 2by – 2 2 2a b y 0

Hence, answer is (a).

Class Exercise - 3

Equation of the circle which passesthrough the origin, has its centre onthe line x + y = 4 and cuts the circlex2 + y2 – 4x + 2y + 4 = 0 orthogonally is

2 2(a) x y 2x 6y 0

2 2(b) x y 6x 3y 0

2 2(c) x y 4x 4y 0

(d) None of these

Solution

Let centre of the circle is (g, 4 – g)[its centre is on x + y = 4]

Equation of circle is 2 2x y 2gx 2 4 g y 0

[ it passes through origin]

Since it cuts the given circle orthogonally,

2 × g × 2 – 2 (4 – g) = 4 6g = 12 g = 2

Equation of the required circle is

2 2x y 4x 4y 0 Hence, answer is (c).

Class Exercise - 4

If O is the origin and OP, OQ aredistinct tangents to the circlex2 + y2 + 2gx +2fy + c = 0, thecircumcentre of the triangle OPQ is

(a) (–g, –f) (b) (g, f)(c) (–f, –g) (d) None of these

Solution

O(0,0)

P

Q

(–g, –f)

PQ is chord of contact of (0, 0) Equation of PQ is gx + fy + c = 0

Family of circles passing through PQand given circle is

2 2x y 2gx 2fy c gx fy c 0

Solution Cont.

It passes through (0, 0) 1

2 2x y gx fy 0

g fCentre is ,

2 2

Hence, answer is (d).

Class Exercise - 5

Prove that the circlex2 + y2 – 6x – 4y + 9 = 0 bisects thecircumference of the circlex2 + y2 – 8x – 6y + 23 = 0.

Solution:

Equation of common chord of the given circle isS1 – S2 = 0

2x 2y 14 0 x y 7 0

Solution Contd..

Centre of circle

is (4, 3) 2 2x y 8x 6y 23 0

which lies on x + y – 7 = 0, first circlebisects the circumference of the secondcircle because common chord passesthrough the end points of a diameterof the second circle.

Class Exercise - 6

If OA and OB are two equal chordsof the circle x2 + y2 – 2x + 4y = 0perpendicular to each other andpassing through the origin then findthe equation of OA and OB.

Solution:

Let chords be y = mx and . Since chords are

of equal lengths, perpendicular distance fromthe centre to the chords will be same.

1

y xm

2 2

1 2m 2 m

1 m 1 m

Solution Contd..

m 2 2m 1

1

m 2 2m 1 m 3,3

Equations of OA and OB are x + 3y = 0 and 3x – y = 0

Class Exercise - 7

The coordinate of two points P and Qare (2, 3) and (3, 2) respectively.If circles are described on OP and OQas diameters, O being the origin thenfind the length of their common chord.

Solution:

(0,0)O

Q(3,2)

P (2,3)

R

ORP , ORQ

2 2

Therefore OR is perpendicularto PQ. We have to find the lengthof OR. From figure it is clear thatOR is length from origin to theline PQ.

Solution Contd..

Equation of PQ is y – 2 = –1 (x – 3)

x y 5 0

5

OR2

Class Exercise - 8

Determine the equation of the circlewhose diameter is the chord x + y = 1of the circle x2 + y2 = 4.

Solution:

Equation of family of circles passingthrough the intersection of circle andline is

2 2x y 4 x y 1 0 ...(i)

2 2x y x y 4 0

Solution Contd..

Its centre lies on x + y – 1 = 0

,2 2

1 0 1

2 2

Substituting value of in the equation (i), we get

2 2x y x y 3 0

Class Exercise - 9

Consider a family of circles passingthrough two fixed points A (3, 7)andB (6, 5). Show that the chords in whichthe circle x2 + y2 – 4x – 6y – 3 = 0cuts the members of the family areconcurrent at a point. Find the coordinateof this point.

Solution:

Family of circles is(x – 3) (x – 6) + (y – 7) (y – 5)

5 7y 7 x 3 0

6 3

x 3 x 6 y 7 y 5 2x 3y 27 0

Solution Contd..

2 2x y 2 9 x 3 12 y 53 27 0

Common chord isS1 – S2 = 0

2 5 x 3 6 y 56 27 0

( 5x 6y 56) (2x 3y 27) 0

This chord is intersection of –5x – 6y + 56 = 0and 2x + 3y – 27 = 0

Solving these equations we get .

232,

3

Class Exercise - 10

If two chords, drawn from the point(p, q) on the circle x2 + y2 = px + qy(where pr 0) are bisected by the x-axis,then

(a) p2 = q2 (b) p2 = 8q2

(c) p2 < 8q2 (d) p2 > 8q2

Solution:

Let the chord is bisected at A (h, 0)P is (p, q)

P Q(p,q) A (h,o)

Solution Contd..

Q (2h – p, –q) lies on circle

2 22h p q p 2h p q q

2 2 2 2 24h p 4ph q 2ph p q

2 2 24h 6ph 2 p q 0

This is quadratic in h and h is real and distinct

D 0

2 2 236p 32 p q 0 2 24p 32q 0

2 2p 8q

Thank you