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Mathematics

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Circle Sessions - 3

Session

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Session Objectives

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Session Objective

1. Equation Of Family Of Circles

2. Equation Of Chord Whosemid Point is Given

3. Equation Of Chord Of Contact

4. Angle Between Two Circles5. Orthogonal Intersection

6. Equation Of Common Chord

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Equation Of Family Of Circles

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

S = 0 L = 0

A

B

S+L = 0

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Questions

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Illustrative Problem

Write the family of circles passing

through the intersection of x2+ y29 =0 and x + y 1 = 0. Find that memberof this family which passes through theorigin.

Solution:

Family of required circle is

S + L = 0

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Solution Cond.

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

Since the required circle passesthrough the origin, we have

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

= -9

Substituting value of in (1) we get

2 2

x y 9x 9y 0

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Equation Of Family Of CirclesPassing 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

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Questions

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Illustrative Problem

Find the equation of the circle passing

through the (4,1), (6,5) and havingits 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

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Solution Cond.

6centre is 5, lies on 4x 3y 16 02

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

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Equation of family of circlewhich 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 thegiven point on it.

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Questions

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Illustrative problem

Find the equation of the circle

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

Solution:

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

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

2 2

x y 25 (3x 4y 25) 0

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Solution Cond.

It passes thorough (1,1)

1 + 1 25 + (3x+4y-25) = 0

23

18

Therefore the required equation ofthe circle is

2 2 23x y 25 (3x 4y 25) 018

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Equation of circles whichtouches a given line at a givenpoint on it

A(x1,y1)

L = 0

2 2

1 1S (x x ) (y y ) 0

family of circle is S + L=0

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Questions

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Illustrative Problem

Find the equation of the circle

passing through origin(0,0) andtouching 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

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Equation of Circle Passing throughthe points of intersection of twocircles

S1+ S2= 0 (-1)

S1= 0S2= 0

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Question

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Illustrative Problem

Find the equation of circle passing

through origin and the points ofintersection 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

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Equation of chord whosemid-point is (x1,y1)

S=0

C=(0,0)

A B(x1,y1)D

xx1+ yy1= x12+ y1

2

T = S1

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Questions

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Illustrative Problem

Find the equation of the chord of

the circle (x-1)2+ (y-2)2= 4 whosemid point is (2,1).

Equation of circle is2 2

x 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 2

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

1T S x y 3 2 x y 1

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Illustrative Problem

Find the equation of the chord of

the circle (x-1)2

+ (y-2)2

= 4 whosemid 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

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

P(x1,y1)

A(x2,y2)

B(x3,y3)

Equation of chord ofcontact is

xx1+yy1=a2

or

T = 0

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Question

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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 :

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Angle at which two circlesintersect

2 2 2

1 2

1 2

d r r

cos 2r r

S1=0 S2= 0

C1 C2

P

r1 r2

d = distance(c1,c2)

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Question

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Illustrative Problem

Find the angle at which the

circles x2+ y28x 2y - 9 =0and x2 + y2 + 2x + 8y - 7 = 0intersect.

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 :

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Orthogonal Intersection

C1C2

r1 r2

S1=0 S2=0

90

d = distance (c1,c2)

Method 1:

d2= r12+r2

2

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Orthogonal Intersection

C1C2

r1 r2

S1=0 S2=0

90

Method 2:

2g1g2+ 2f1f2= c1+ c2

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Question

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Illustrative Problem

If two circles of equal radii a withcentre (2,3) and (5,6) respectively cuteach other orthogonally then find thevalue of a.

Two circles cut orthogonally

2 2 2

1 2d r r

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

Therefore a2 + a2 = 18 => a = 3

Solution :

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Common Chord Of two circlesS1 = 0 and S2= 0

S1=0 S2=0

A

B

Equation of common chord is S1 - S2= 0

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Question

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Illustrative Problem

Find the equation of common chord of

two circles x2+ y2=25 and4x2 + 4y2 - 40x + 91=0

2 2

1here S x y 25 0 2 2

2

91S x y 10x 0

4

1 2S S 0 40x 191 0

Solution :

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Class Test

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Class Exercise - 1

A variable chord is drawn through the

origin 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

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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).

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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

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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).

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Class Exercise - 3

Equation of the circle which passes

through the origin, has its centre onthe line x + y = 4 and cuts the circlex2+ y24x + 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

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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).

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Class Exercise - 4

If O is the origin and OP, OQ are

distinct 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

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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

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Solution Cont.

It passes through (0, 0) 1

2 2x y gx fy 0

g fCentre is ,

2 2

Hence, answer is (d).

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Class Exercise - 5

Prove that the circle

x2

+ y2

6x 4y + 9 = 0 bisects thecircumference of the circlex2+ y28x 6y + 23 = 0.

Solution:

Equation of common chord of the given circle isS1S2= 0

2x 2y 14 0 x y 7 0

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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 second

circle because common chord passesthrough the end points of a diameterof the second circle.

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Class Exercise - 6

If OA and OB are two equal chordsof the circle x2+ y22x + 4y = 0

perpendicular 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

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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

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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 , ORQ2 2

Therefore OR is perpendicularto PQ. We have to find the length

of OR. From figure it is clear thatOR is length from origin to theline PQ.

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Solution Contd..

Equation of PQ is y 2 = 1 (x 3)

x y 5 0

5

OR2

l

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Class Exercise - 8

Determine the equation of the circlewhose diameter is the chord x + y = 1

of 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

S l i C d

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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

Cl E i 9

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Class Exercise - 9

Consider a family of circles passingthrough two fixed points A (3, 7)and

B (6, 5). Show that the chords in whichthe circle x2 + y24x 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

S l ti C td

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Solution Contd..

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

Common chord isS1S2= 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

Cl E i 10

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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) p

2< 8q2 (d) p2> 8q2

Solution:

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

P is (p, q)

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

S l ti C td

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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

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Thank you