SYSTEM OF CIRCLES MATHEMATICS · Note : The angle between two circles at either point of intersection is the same. THEOREM 2.1 Let C 1, C 2 be the centres of two intersecting circles

1

SYSTEM OF CIRCLES MATHEMATICS

NARAYANA IIT CO SPARK PROGRAM (HYD-MDPCOS)

Proof :Let P,Q be the points of intersection of the two

circles. Let the tangents drawn to two circles at Pintersect the line joining the centres 1 2C C

at T1 and

T2. Then 1 2T PT

P

C1 C2

QFig 2.2

T1 T2

Now, 1 2 1 2 2 2C PC C PT T PC

0 0 090 90 180

From 1 2 ,C PC

2 2 21 2 1 2 1 2 1 22 cosC C C P C P C P C P C PC

i.e., 2 2 2 01 2 1 22 cos(180 )d r r r r

2 2 21 2

1 2cos

2d r r

r r

Since cos is independent of the point ofintersection (i.e., coordinates of the point ofintersection are not involved), the angle at Q is alsoequal to

Note : Let the angle between the circles at Q be .

Then 1 2 180C QC

C1P = C1Q = r1, C2P = C2Q = r2

The triangles C1PC2, C2QC2 are identical

1 2 1 2C PC C QC 0 0180 180

INTRODUCTION

In this chapter, we shall discuss the anglebetween two intersecting circles and obtain thecondition for their orthogonality. We shall alsodiscuss about the radical axis of two circles andradical centre of three circles. Further, we alsodiscuss about the coaxal system of circles andsystem of circles orthogonal to a given coaxal systemof circles.

2.1 ANGLE BETWEEN TWOINTERSECTING CIRCLES

Definition :

The angle between two intersecting circlesis defined as the angle between the tangents atthe point of intersection of the two circles.

P

T1 T2

C1 C2

QFig 2.1

1 2T PT is the angle between the circles at P.

Note : The angle between two circles at either point ofintersection is the same.

THEOREM 2.1

Let C1, C2 be the centres of two intersectingcircles of radii r1 and r2 . Let C1C2=d. If is theangle between the circles then

2 2 21 2

1 22d r rCos

r r

SZ1-A DAILY PRACTICE PAPER MATHEMATICSTOPIC : SYSTEM OF CIRCLES DATE: 30-04-2020


2 NARAYANA IIT CO SPARK PROGRAM (HYD-MDPCOS)

THEOREM 2.2

If is the angle between the intersectingcircles x2 + y2 + 2gx + 2fy + c = 0 and x2 + y2 +2g'x + 2f'y + c' = 0, then

2 2 2 2

2 2cos2

c c gg ff

g f c g f c

Proof:

Let C1, C2 be the centres and r1, r2 be theradii of the two given circles.

C1 = (–g, –f), C2 = (–g', –f')

2 2 2 21 2,r g f c r g f c

2 2 21 2 1 2

1 2cos

2c c r r

r r

2 2 2 2 2 2

2 2 2 22

g g f f g f c g f c

g f c g f c

2 2 2 2

2 2cos2

c c gg ff

g f c g f c

Ex: Find the angle between the circles given by theequations x2 + y2 + 6x – 10y – 135 = 0,x2 + y2 – 4x + 14y – 116 = 0 .

Sol: Given circles are

x2 + y2 + 6x – 10y – 135 = 0 -- (1)

x2 + y2 – 4x + 14y – 116 = 0 -- (2)

Hence c1 = (–3, 5), 1 9 25 135 13r

c2 = (2, –7), 2 4 49 116 13r

1 2 25 144 13d c c

Let be the angle between (1) & (2)

2 2 2

1 2

1 2

169 169 169 1cos2 2 169 2

d r rr r

23

2.2 ORTHOGONAL CIRCLES

Definition :

If the angle between two circles is a rightangle, then the two circles are said to cut eachother orthogonally.

THEOREM 2.3

Let d be the distance between the centresof two inter sect ing cir cles of r adi i r1 and r2. Ifthe two circles cut each other orthogonally, then

2 2 21 2d r r .

Proof :

Let be the angle between the circles

The circles are orthogonal 2

2 2 21 2cos cos 0 0

2d r r

2 2 21 2d r r

2 2 21 22

d r r

THEOREM 2.4

The condition that the two circles S = 0 andS' = 0 may cut each other orthogonally is 2gg' +2ff' = c + c'.

Proof :

Let the circlesS = x2 + y2 + 2gx + 2fy + c = 0S' = x2 + y2 + 2g'x + 2f'y + c' = 0intersect orthogonally at P

C1 = (–g, –f) and C2 = (–g', –f') are the centresof the circles whose radii are respectively

2 21r g f c and 2 2

2r g f c

2r

2C1C

P0901r

2.3Fig

3



From the right angled triangle C1PC2,2 2 2 2 2

1 2 1 2 1 2C C C P C P r r

(–g + g')2 + (–f + f')2

= g2 + f2 – c + g'2 + f'2 – c'

2gg' + 2ff' = c + c'

Note : In the case of orthogonal circles, the tangent ofone of the circles at the point of intersection willbe a normal to the other circle and hence it passesthrough the centre of the other circle.

Ex: Find k if the circles x2 + y2 – 5x – 14y – 34 = 0 andx2 + y2 + 2x + 4y + k = 0 are orthogonal to each other.


x2 + y2 – 5x – 14y – 34 = 0 - - - (1)

x2 + y2 + 2x + 4y + k = 0 - - - (2)

(1) & (2) are orthogonal to each other

2gg' + 2ff' = c + c'

52 1 2 7 2 342

k

– 5 – 28 = –34 + k

k = 1

THEOREM 2.5

(i) If S = 0, S' = 0 are two circles intersectingat two distinct points, then S – S' = 0 (or S' – S =0) represents the common chord of these circles.

(ii) If S = 0, S' = 0 are two circles touching eachother, then S – S' = 0 (or S' – S = 0) is a commontangent at the point of contact of the two circles.

Proof :

Let S = x2 + y2 + 2gx + 2fy + c = 0 -- (1)

and S' = x2 + y2 + 2g'x + 2f'y + c = 0 -- (2)

(i) Let P(x1, y1) and Q(x2, y2) be the points ofintersection of (1) & (2).

Consider S – S' = 0

2(g – g') x + 2 (f – f') y + (c – c') = 0 -- (3)

Clearly the points P, Q lie on (3) since1 1

11 22 11 220, 0, 0, 0S S S S .

Equation (3) a linear in x and y and hence itrepresents a line.

S – S' = 0 is the equation of the commonchord of (1) & (2).

1C 2C

A

B

2.4Fig

0S 0S

1C 2CP

0S 0S

2.5Fig

(ii) Let (1) & (2) touch each other at P(x1, y1)Consider S – S' = 0

2(g – g') x + 2(f – f') y + (c – c') = 0 --- (4)P(x1, y1) is a point on (4) and it represents a

line and the slope of (4) is .g gf f

Slope of the line joining the centres of the

circles is f fg g .

1C 2C P

2.6Fig

S

S

The line (4) is perpendicular to the line ofcentres and it passes through the point of contact ofthe two circles. Hence it is a tangent common toboth the circles.

THEOREM 2.6

(i) I f a cir cle S x2 + y2 + 2gx + 2fy + c = 0 anda line L lx + my + n = 0 intersect each otherthen the equation of the circle passing throughthe points of intersection of the circle and theline is given by S + kL = 0 where k is a realparameter.

(ii) If the circles S x2 + y2 + 2gx + 2fy + c = 0and S' x2 + y2 + 2g'x + 2f'y + c' = 0 intersecteach other then the equation of the circle passingthrough the points of intersection of S = 0 andS' = 0 is given by 0S S where and are parameters such that 0 .



Proof :(i) Consider the equation

(x2 + y2 + 2gx + 2fy + c) + k(lx + my + n) = 0i.e., S + KL = 0 -- (1)Clearly, this equation represents a circle

0S 0L

0S kL

P

Q

2.7Fig

In order to prove that (1) represents a circlepassing through the points of intersection of thecircle S = 0 and the line L = 0, it is sufficient toshow that their points of intersection satisfy equation(1)

Let P(x1, y1) and Q(x2, y2) be the points ofintersection of S = 0 and L = 0.

Then 2 21 1 1 12 2 0x y gx fy c

2 22 2 2 22 2 0x y gx fy c

lx1 + my1 + n = 0lx2 + my2 + n = 0

2 21 1 1 1( 2 2 )x y gx fy c

1 1( ) 0k x my n -- (2)

and 2 22 2 2 2( 2 2 )x y gx fy c

2 2( ) 0k x my n -- (3)

From (2) & (3) it follows that the points P andQ satisfy the equation (1).

(1) represents a circle passing through thepoints of intersection of S = 0 & L = 0.

(ii) Consider 0S S ------- (4)

where , are any real numbers such that0 . Equation (4) clearly represents a circle

and it passes through the points of intersection ofS = 0 and S' = 0.

0S P

Q

2.8Fig

0S

0S S

Further, equation (4) is equivalent to S+KL=0

where

k and L S – S' = 0.

Note 1: The equation 0S S can also be written as

S + kS' = 0 . where 1k

(assuming 0 )

Note 2: If k = –1, then S + kS' = S – S' = 0 represents thecommon chord of the circles S = 0 and S' = 0. If thepoints of intersection coincide, then S – S' = 0 is acommon tangent to the circles.

SOLVED EXAMPLES

1. If the angle between the circles

x2 + y2 – 12x – 6y + 41 = 0 and

x2 + y2 + kx + 6y – 59 = 0 is 450 find k.

Sol: Here g = –6, f = –3, c = 41,

, 3, 592kg f c

Given 045

2 2 2 2

2 2cos2

c c gg ff

g f c g f c

0

2

41 59 2 6 2 3 32cos 45

2 36 9 41 9 594

k

k

2

314

22 68

4

kk

k

2. If 2 2 6 8 0x y x y k and 2 2 4 6 24 0x y x y are orthogonal then

find k.

5



Sol: Given circles x2 + y2 – 6x – 8y + K = 0 ... (1)

x2 + y2 – 4x – 6y + 24 = 0 .... (2)

(1) & (2) are cuts each other orthogonally

2g.2g1 + 2f.2f1 = 2(c + c1)

24 + 48 = 2(K + 24)

K = 12

3. Find the equation of the circle which passes throughthe point (0, –3) and intersect the circles given by theequations x2 + y2 – 6x + 3y + 5 = 0 andx2 + y2 – x – 7y = 0 orthogonally.

Sol: Let A = (0, –3)

Given circles are x2 + y2 – 6x + 3y + 5 = 0 -- (1)

x2 + y2 – x – 7y = 0 -- (2)

Let the equation of the required circle be

x2 + y2 + 2gx + 2fy + c = 0 -- (3)

(3) cuts (1) orthogonally

32 3 2 52

g f c

–6g + 3f – c = 5 -- (4)

(3) cuts (2) orthogonally

1 72 22 2

g f c

–g – 7f – c = 0 -- (5)

(3) Passes through 0 9 0 6 0A f c

6 9f c -- (6)

Solving (4), (5) and (6) we get

1 2,3 3

g f and 5c

Equation of the required circle is

2 2 2 45 0

3 3x y x y

2 23 2 4 15 0x y x y

4. If P and Q are conjugate points w.r.t a circleS = x2 + y2 + 2gx + 2fy + c = 0, then prove that thecircle PQ as diameter cuts the circle S = 0orthogonally.

Sol: Let P = (x1, y1) and Q = (x2, y2)S = x2 + y2 + 2gx + 2fy + c = 0

The polar of P w.r.t S = 0 is

1 1 1 1 0 xx yy g x x f y y c

P and Q are conjugate w.r.t 120 0 S S

1 2 1 2 1 2 1 2 0 x x y y g x x f y y c -- (1)

Equation of the circle with PQ as diameter is

1 2 1 2 0 S x x x x y y y y

2 21 2 1 2 S x y x x x y y y

1 2 1 2 0 x x y y

1 22 g x x and 1 22 f y y ,

1 2 1 2 c x x y y

1 2 1 22 2 gg ff g x x f y y

1 2 1 2 c x x y y c c

The circles S = 0 and S' = 0 cut each otherorthogonally.

5. If the equations of two circles whose radii are a anda' are S = 0 and S' = 0, then show that the circles

0S Sa a

and 0

S Sa a

intersect orthogonally..

Sol: With out loss of generality, let the line of centres bechoosen as x - axis and the distance between them be2p , origin being the mid - point.

The centres of the circles are (p, 0) and (–p, 0)

2 2 2 S x p y a

2 2 2 22 0 S x y px p a

2 2 2 S x p y a

2 2 2 22 0 S x y px p a

Circles are 0 0

S S

a S aSa a

' ' 0 a S aS

2 2 2

2 0

a a x y p

px a a aa a a



2 2 22 0 a ax y p x p aaa a

0 a S aS

2 2 22 0 a ax y p x p aaa a

2 2 2 0

a a a agg ff p pa a a a

22 p c c

The circles 0

S Sa a

cut each other

orthogonally

6. Find the equation of the circle passing through thepoints of intersection of circlesx2 + y2 + 6x + 4y – 12 = 0, x2 + y2 – 4x – 6y – 12 = 0, andhaving radius 13 .

Sol: Given circles

S x2 + y2 + 6x + 4y – 12 = 0

S1 x2 + y2 – 4x – 6y – 12 = 0

R.A. of S = 0 and S1 = 0 is S – S1 = 0

10x + 10y = 0

0 L x y

The equation of the circle passing through the pointof intersection of S = 0 and L = 0 is

2 0S L

x2 + y2 + 6x + 4y – 12 + 2 (x + y) = 0

x2 + y2 + 2x (3 ) + 2y (2 ) –12 = 0 .... (1)

Given radius = 13

2 2(3 ) (2 ) 12 13

22 10 13 12 13

2 5 6 0

2( ) 3 or

required circles, from (1)

x2 + y2 + 2x – 12 = 0,

x2 + y2 – 2y – 12 = 0

7. Find the equation of the circle which cutsorthogonally the circle 2 2 6 4 12 0x y x y and having the centre at (–1, 2).

Sol: Given circles 2 2 6 4 12 0 S x y x y

Let the equation of the circle cuts S = 0 orthogonallyis

1 2 2 2 2 0 S x y gx fy c

Given centre of S1 = 0 is (–1, 2)

(–g, –f) = (–1, 2) g = 1, f = –2

from S = 0, S1 = 0

2g (–6) + 2f(4) = 2(c – 12)

–6g + 4f = c –12

c = –2

1 2 2 2 4 2 0 S x y x y

8. Find the equation of the circle passing through theintersection of the circles x 2 + y2 = 2ax, x2 + y2 = 2by

and having its centre on the line 2x ya b .

Sol: Given circles are 2 2 2 0 S x y ax -- (1) and 2 2 2 0 S x y by -- (2)

The equation of the common chord of (1) & (2) is0 0L S S L ax by -- (3)

The equation of the circle passing through theintersection of (1) & (2) is

0S L 2 2 2 0x y ax ax by

2 2 2 0x y ax by -- (4)

is a parameter..

Centre of (4), 2,

2 2a bC

Since C lies on 2x ya b , 2

22 2

a ba b

2 4 2 2 1

Equation of the required circle is2 2 3 0x y ax by

7



EXERCISE – 2.1

SHORT ANSWER QUESTIONS

1. Find the acute angle between the followingintersecting circles.i) 2 2 12 6 41 0x y x y

2 2 4 6 59 0x y x y ii) 2 2 4 14 28 0x y x y

2 2 4 5 0x y x (June 2002)

2. Show that the angle between the circles2 2 2x y a , 2 2x y ax ay is 3

4

3. Show that the circles given by the followingequations intersect each other orthogonally

i) 2 2 2 0x y my g 2 2 2 0x y x g

ii) 2 23 3 8 29 0x y x y 2 2 2 2 7 0x y x y

iii) 2 2 4 2 11 0x y x y 2 2 4 8 11 0x y x y

iv) 2 2

2 22 4 4 0,3 4 1 0

x y x yx y x y

4. Find k if the following pairs of circles areorthogonali) 2 2 2 0x y by k

2 2 2 8 0x y ax

ii) 2 2

2 26 8 12 0,4 6 0

x y x yx y x y k

iii) 2 2 16 0x y y k 2 2 4 8 0x y x

5. Find the equation of the circle whichcuts orthogonally the circle

2 2 4 2 7 0x y x y and having thecentre at (2, 3).

9. If the straight line represented bycos sinx y p intersect the circle 2 2 2x y a

at the points A and B, then show that the equation ofthe circle with AB as diameter is

2 2 2 2 cos sin 0x y a p x y p

Sol: The equation of the circle passing through the pointsA and B is 2 2 2 cos sin 0x y a x y p -- (1)

is a parameter

Centre of (1), cos sin,

2 2C

Since the line cos sin 0x y p is a diameterof (1) C lies on the line

cos sincos sin

2 2p

2p


2 2 2 2 cos sin 0x y a p x y p

10. Find the equations of the circles which cutorthogonally the circles 2 2 6 1 0x y y ,

2 2 4 1 0x y y and touch the line3 4 5 0x y

Sol: Let the required circle be 2 2 2 2 0x y gx fy c

-- (1)Given circles are 2 2 6 1 0x y y -- (2)

2 2 4 1 0x y y -- (3)(1) cuts (2) and (3) orthogonally

2 0 2 3 1g f c 6 1f c

and 2 0 2 2 1g f c 4 1f c

0f and 1c (1) Touches the line 3 4 5 0x y

2 23 4 5

5g f

g f c

23 51

5g

g

2 23 5 25 1g g

2 1516 30 0 0,8

g g g

The equation of the required circles are

2 2 1 0x y and 2 24 15 4 0x y x



LONG ANSWER QUESTIONS1. Find the equation of the circle which passes

through the origin and intersects each of thefollowing circles orthogonally. i) 2 2 4 6 10 0x y x y , 2 2 12 6 0x y y

ii) 2 2

2 24 6 3 0,8 12 0

x y x yx y y

2. Find the equat ion of the circle whichpasses through (1, 1) and cuts orthogonally eachof the circles 2 2 8 2 16 0x y x y and

2 2 4 4 1 0x y x y .

3. Find the equation of the circle which passthrough the points (2, 0) (0, 2) and orthogonalto the circle 2 22 2 5 6 4 0x y x y .

4. Find the equation of the circle passingthrough the origin, having its centre onthe line x + y = 4 and intersecting the circle

2 2 4 2 4 0x y x y orthogonally..(April 2001)

5. Find the equation of the circle which cutsthe circles 2 2 4 6 11 0x y x y and

2 2 10 4 21 0x y x y orthogonally andhas the diameter along the line 2 3 7x y .

6. Find the equation of circle which intersect thecircle 2 2 6 4 3 0x y x y orthogonallyand passes through the point (3, 0) and touchesY-axis.

7. Find the equation of the circle passing throughthe points of intersections of the circles

2 2 8 6 21 0x y x y ,2 2 2 15 0x y x and (1, 2).

8. i) If the straight line 2x + 3y = 1 intersectsthe circle x2 + y2 = 4 at the points A and B,then find the equation of the circle havingAB as diameter.

ii) If x + y = 3 is the equation of the chord ABof the circle 2 2 2 4 8 0x y x y ,

find the equation of the circle having ABas diameter.

ANSWERS


1. i)4

ii) 0603

4. i) k = 8 ii) k = –24 iii) 8k 5. 2 2 4 6 9 0x y x y

LONG ANSWER QUESTIONS

1. i) 2 22( ) 7 2 0x y x y

ii) 2 2 6 3 0x y x y

2. 2 23( ) 14 23 15 0x y x y

3. 2 27 7 8 8 12 0x y x y

4. 2 2 4 4 0x y x y

5. 2 2 4 2 3 0x y x y

6. 2 2 6 6 9 0x y x y

7. 2 23 18 12 27 0x y x y

8. i) 2 213 4 6 50 0x y x y

ii) 2 2 6 4 0x y x

2.3 RADICAL AXIS OF TWO CIRCLES

Definition :The locus of a point, the powers of which

with respect to two given circles are equal, is astraight line, called the radical axis of the twocircles.

THEOREM 2.7

The radical axis of two circles S = 0and S' = 0 is the straight line S – S' = 0.

Proof:

Let 2 2 2 2 0S x y gx fy c and2 2 2 2 0S x y g x f y c

Let P(x1, y1) be any point on the radical axis.Then the powers of P with respect to the circles areequal

9



11 11 11 11 0S S S S

1 12 2 0g g x f f y c c

Locus of P(x1, y1) is

2 2 0g g x f f y c c )

i.e., S – S' = 0.

The equation of the radical axis of thecircles S = 0 & S' = 0 is S – S' = 0.

Equation (1) is a first degree equation in x andy, so it represents a straight line

Radical axis of two circles is a straight line.Note 1 : For concentric circles, the radical axis doesnot

exist, since there is no point whose powers withrespect to two distinct concentric circles are equal.(However, if their radii are equal, then the locus isthe whole plane). So whenever we consider theradical axis of two circles, it means that two circlesare non - concentric.

Note 2: While using the formula S–S' = 0 to find theequation of the radical axis, the equations of thecircles must be in the general form.

Note 3: The lengths of the tangents, drawn to two circlesfrom any point (outside the two circles) on theirradical axis, are equal.

THOEREM 2.8

The radical axis of any two circles isperpendicular to their line of centres.Proof :

Let the circles be2 2 2 2 0S x y gx fy c 2 2 2 2 0S x y g x f y c

Hence 1 2, & ,C g f C g f

Equation of the radical axis is S–S' = 0

2 2 0g g x f f y c c -(1)

Slope of (1) =

g gf f

Slope of 1 2f f

C Cg g

Since the product of the slopes is –1, the

radical axis is perpendicular to the line of centres.

THEOREM 2.9The radical axis of any two intersectingcircles is their common chord.

1C 2C

A

B

0S 0S

2.9FigProof :Let the two circles S = 0 and S' = 0 intersect

at A(x1, y1) and B(x2, y2)Since A(x1, y1) lies on both the circles, then

S11= 0 and S'11= 0 11 11 0S S -- (1)

Similarly B(x2, y2) lies on both the circles, thenS22= 0 and S'22= 0 22 22 0S S -- (2)

Hence A(x1, y1) and B(x2, y2) satisfy theequation S–S' = 0, which is the radical axis of twocircles.

The common chord AB (produced) is theradical axis of the circles. THEOREM 2.10

The radical axis of two circles touching eachother is the common tangent at their point ofcontact.

1C 2CP

0S 0S

2.10Fig

1C 2C P

S

S

2.11Fig

Proof :Let P(x1, y1) be the point of contact of the two

circles S = 0 and S' = 0 touching each other.

11 0S and 11 0S 11 11 0S S Hence the point P(x1, y1) lies on the radical

axis S – S' = 0.The radical axis passing through the point of

contact also being perpendicular to the line ofcentres will be the common tangent.



THEOREM 2.11The radical axes of three circles whose

centres are non - collinear, taken in pairs, areconcurrent.

P

1L

2L3L

0S

0S

0S 2.12Fig

Proof :Let the equations of three circles (whose

centres are not collinear) be2 2 2 2 0S x y gx fy c -- (1)2 2 2 2 0S x y g x f y c -- (2)2 2 2 2 0S x y g x f y c -- (3)

The radical axis L1(say) of (1) and (2) is 1 2 2 0L g g x f f y c c

-- (4) Similarly, the radical axis L2 (say) of (2)&(3) is 2 2 2 0L g g x f f y c c

-- (5)and the radical axis L3 (say) of (3) & (1) is

3 2 2 0L g g x f f y c c -- (6)

Now L1+L2+L3 = 0 shows that L1, L2 and L3are concurrent.

2.4 RADICAL CENTREDefinition :

The point of concurrence of the radicalaxes, of three circles whose centres are non -collinear, taken in pairs, is called the radicalcentre of the three circles.Note : The powers of the radical centre w.r.t each of the

three circles are equal.

THEOREM 2.12The radical axis of the two circles (whose

common tangent is not parallel to the line joiningthe points of contact of common tangent to thecircles) bisects the line joining the points ofcontact of common tangent to the circles.Proof :

Let 2 2 2 2 0S x y gx fy c -- (1)2 2 2 2 0S x y g x f y c -- (2)

be two circles and T1, T2 be the points ofcontact of common tangent to the circles S = 0 &S' = 0.

We know that radical axis of two circles isperpendicular to the line joining the centres of circles.The common tangent is not perpendicular to the linejoining the centres. Therefore common tangent andradical axis intersect at a point.

Let T1T2 (common tangent) intersect theradical axis of (1) and (2) at P(x1, y1).

The powers of P with respect to the circlesS = 0 and S' = 0 are equal.

1 1 2 2. .PT PT PT PT

2 21 2 1 2PT PT PT PT

i.e., P is the mid point of T1 and T2.The radical axis of the two circles bisects

each of their common tangents.

THEOREM 2.13If a circle cuts two given circles

orthogonally then its centre lies on the radicalaxis of the circlesProof:-

Let x2 +y2 + 2gx + 2fy +c = 0 be the circlewhich cuts the given circles say

x2+y2+2gix+2fiy+ci= 0,i = 1, 2 orthogonallyThen 2gg1 + 2ff1 = c + c1 ------- (1)

and 2gg2 +2ff2 = c + c2 ------- (2)

from (1), (2)

2g(g1–g2) + 2f (f1 – f2) = c1 – c2

11



i.e., 2 0 0g f c c

0 0 Similarly, the condition of orthogonality is

satisfied by (2) and (4), and (3) and (4).

The circle having radical centre of threecircles as the centre and having the length of tangentfrom the radical centre to any one of these circles asradius cuts the given three circles orthogonally.

SOLVED EXAMPLES

1. Find the equation of the radical axis of the circlesrepresented by 2 22 2 3 6 5 0x y x y and

2 23 3 7 8 11 0x y x y

Sol: The given circles are

2 2 3 53 02 2

S x y x y -- (1)

and 2 2 7 8 11 03 3 3

S x y x y -- (2)

Radical axis of (1) & (2) is S – S' = 0

3 7 8 5 113 02 3 3 2 3

x y

23 2 7 0x y 2. Find the radical centre of the circles

x2 + y2 + 4x – 7 = 0, 2x2 + 2y2 + 3x + 5y – 9 = 0 andx2 + y2 + y = 0.

Sol: Given circles are 2 2 4 7 0S x y x -- ( 1)

2 2 3 5 90

2 2 2S x y x y -- (2)

2 2 0S x y y -- (3)

Radical axis of (1) & (2) is S – S' = 0

3 5 94 7 02 2 2

x y

5 5 5 02 2 2

x y

1 0x y -- (4)

Radical axis of (1) & (3) is 0S S

4 7 0x y -- (5)

Solving (4) & (5), we get

i.e 2(g1–g2) (–g) + 2(f1 – f2) (–f) + c1– c2 = 0 -------(3)

Since the equation of radical axis of the given circles is 2(g1–g2)x + 2(f1– f2)y + c1–c2 = 0,clearly the point (–g,–f) lies on it from (3)

THEOREM 2.14

Let S' = 0, S''= 0, S'''= 0 be three circleswhose centres are non-collinear and no twocircles of these are intersecting, then the circlehaving

(i) radical centre of these circles as the centre.

(ii) length of tangent from the radical centreto any one of these three circles as radius cutsthe given three circles orthogonally.Proof :

Let C be the radical centre of three givencircles. As no two of the three circles areintersecting, the point C is exterior to these circles.Choose C as the origin.

Let the equations of the three circles be2 2 2 2 0S x y g x f y c -- (1)2 2 2 2 0S x y g x f y c -- (2)2 2 2 2 0S x y g x f y c -- (3)

The length of the tangent from C to (1) is c .Since the origin is an external point to these circles,we have c', c'', c''' are positive.

Now, the equation of the circle having thecentre as C and radius c is 2 2 0x y c -- (4)

Now, we prove that (4) is orthogonal to (1),(2) and (3). The lengths of the tangents from C to(2) & (3) are ,c c respectively..

The lengths of the tangents frsom the radicalcentre to these three circles are equal.

c c c c c c The circles (1) & (4) are orthogonal since the

condition of orthogonally is satisfied.



Radical centre of the circles = (2, 1)

3. Show that the common chord of the circles2 2 6 4 9 0x y x y and x2 + y2 – 8x–6y+23= 0

is the diameter of the second circle and also find itslength.

Sol: Given circles are2 2 6 4 9 0S x y x y -- (1)2 2 8 6 23 0S x y x y -- (2)

The equation of the common chord of (1) & (2) is

0 2 2 14 0S S x y 7 0x y -- (3)

Centre of (2), C = (4, 3)

Clearly C satisifies (3)

(3) is a diameter of (2)

Length of common chord = Diameter of (2)

2 16 9 23 2 2

4. Prove that the radical axis of the circles x2 + y2 +2gx + 2fy + c = 0 and x2 + y2 + 2g'x + 2f'y + c = 0 isthe diameter of the later circle (or the formerbisects the circumference of the later) if2g'(g–g')+2f'(f–f') = c–c'


2 2 2 2 0S x y gx fy c -- (1)

2 2 2 2 0S x y g x f y c -- (2)

The radical axis of (1) & (2) is S–S' = 0

2 2 0g g x f f y c c -- (3)

Centre of (2), ,C g f

Radical axis is the diameter of (2) if C lies on (3)

i.e., if 2 2 0g g g f f f c c

i.e., if 2 2g g g f f f c c ,

which is true.Note : A circle S = 0 is said to bisect the circumference of

a circle S' = 0 if the common chord of the two circlespasses through the centre of S' = 0 i.e., if thecommon chord is a diameter of S' = 0.

5. Find the equation of the circle which cuts the circles2 2 4 2 1 0x y x y , 2 22( ) 8 6 3 0x y x y

and 2 2 6 2 3 0x y x y orthogonally..

Sol: Method I :

Given circles are2 2 4 2 1 0S x y x y -- (1)

2 2 34 3 0

2S x y x y -- (2)

2 2 6 2 3 0S x y x y -- (3)

Equation of the radical axis of (1) & (2) is

5 50 0

2 2S S y y -- (4)

Equation of the radical axis of (1) & (3) is

0 2 4 4 0S S x y

2 2 0x y -- (5)

Solving (4) & (5), we get x = 7

Radical centre of (1), (2) & (3) is 57,2

C

Length of the tangent from C to (1) = 11S

2549 28 5 14

357 357

4 2

Equation of the circle which is orthogonal to

(1), (2) & (3) is 2

2 5 35772 4

x y 2 2 14 5 34 0x y x y

Method II :

Given circles 1 2 2 4 2 1 0 S x y x y

11 2 2 4 3 3 / 2 0 S x y x y

111 2 2 6 2 3 0 S x y x y

Let the equation of the required circle2 2 2 2 0S x y gx fy c

S = 0, S1 = 0 cuts orthogonally

2 (4) 2 (2) 2( 1)g f c

4g + 2f = c + 1 ............ (1)S = 0, S11 = 0 cuts orthogonally2g(4) + 2f(3) = 2(–3/2 + c)

34 3

2

g f c .......... (2)

S = 0, S111 = 0 cuts orthogonally

2g(6) + 2f(–2) = 2(c – 3)

13



6g – 2f = c – 3 ........ (3)

(1) – (2) – f = 5/2 f = –5/2

(1) – (3) – 2g + 4f = 4

–2g – 10 = 4

g = –7

g = 3, 52

f

From equation (1)

4(–7) + 2(–5/2) = c + 1 c = –34

Required circle is 2 2 14 5 34 0S x y x y

EXERCISE 2.2

VERY SHORT ANSWER QUESTIONS1. Find the equation of the radical axis of the

following circles.i) 2 2 2 4 1 0x y x y ,

2 2 4 6 5 0x y x y ii) 2 2 3 4 5 0x y x y

2 23( ) 7 8 11 0 x y x yiii) 2 2 4 6 7 0x y x y

2 24( ) 8 12 9 0x y x y 2. Find the equation of the common chord of the

following pair of circlesi) 2 2 2 3 1 0x y x y

2 2 4 3 2 0x y x y ii) 2 2 2x a y b c

2 2 2x b y a c a b

3. Find the equation of the common tangent ofthe following circles at their point of contact.

i) 2 2 10 2 22 0x y x y , 2 2 2 8 8 0x y x y

ii) 2 2 8 4 0x y y , 2 2 2 4 0x y x y

iii) x2+y2–4x+6y+8=0,

x2+y2–10x–6y+14= 0.

4. If x – y – k = 0 is the radical axis of x2+y2+10x–4y–1=0 and x2+y2+5x+y+4=0 then find k.

SHORT ANSWER QUESTIONS1. Find the radical centre of the following circles

i) 2 2 4 6 5 0x y x y 2 2 2 4 1 0x y x y 2 2 6 2 0x y x y

ii) 2 2 2 6 0x y x y 2 2 4 2 6 0x y x y 2 2 4 2 6 0x y x y

2. Find the area of the triangle formed by thecommon chord of two circles 2 2 2 4 11 0S x y x y and 2 2 4 6 4 0'S x y x y and the coor-

dinate axes.

3. Find the condit ion that the circle 2 2

1 1 12 2 0x y g x f y c bisects the cir-cumference of the circle

2 22 2 22 2 0x y g x f y c .

4. Show that the circle 2 2 4 4 1 0x y x y bisects the circum-

ference of the circle 2 2 2 3 0x y x .

LONG ANSWER QUESTIONS1. Find the equation and length of the common

chord of the circles 2 2 5 6 4 0x y x y ,2 2 2 2 0x y x

2. Find the equation of the circle whosediameter is the common chord of the circlesx2+ y2 + 2x + y2 + 2x + 3y +1 = 0 andx2 + y2 + 4x + 3y + 2 = 0

3. Find the equation of the circle which cuts eachof the following circles orthogonallyi) 2 2 3 2 1 0x y x y

2 2 6 5 0x y x y 2 2 5 8 15 0x y x y (March 2007)



ii) 2 2 2 4 1 0x y x y 2 22 2 6 8 3 0x y x y

2 2 2 6 3 0x y x y

iii) 2 2 2 17 4 0x y x y 2 2 7 6 11 0x y x y 2 2 22 3 0x y x y

iv) 2 2 0x y x y 2 22 2 5 3 2 0x y x y

2 2 4 2 2 0x y x y ANSWERS

VERY SHORT ANSWER QUESTIONS

1. i) 3 0x y ii) 10 2 0x y

iii) 8x + 12y + 9 = 0

2. i) 2 1 0x y ii) x – y = 0

3. i) 4 3 7 0x y ii) 2 2 0x y

4. –1


1. i)7 11,6 6

ii)

30,4

2. 15/8 sq. units

3. 2 1 2 2 1 2 1 22 ( ) 2 ( )g g g f f f c c


1. 2 2 0x y , 1425

2. 2 22 2 6 1 0x y x y

3. i) 2 2 6 4 14 0x y x y

ii) 2 2 5 14 34 0x y x y

iii) 2 2 6 4 44 0x y x y

iv) 2 2( 1) ( 1) 2x y

2.5 COAXAL SYSTEM OF CIRCLES

Definition :A system of circles in which every pair of

circles has the same radial axis is called a systemof coaxal circles or a coaxal system.

Since the radical axis is perpendicular to theline of centres, it follows that the centres of thecircles forming a coaxal system are collinear.

THEOREM 2.15

If S = 0, S'= 0 are two circles, then1 2 0S S ( 1 , 2 are parameters not

both zero and 1 2 0 ) represents a coaxalsystem of circles of which S = 0, S'= 0 are itsmembers.Proof :

Let the given circles be2 2 2 2 0S x y gx fy c -- (1)

and 2 2 2 2 0S x y g x f y c -- (2)

Consider the equation 1 2 0S S . 2 2

1 2 1 2 1 22x y x g g 1 2 1 22 0y f f c c -- (3)

In this equation, the coefficients of x2 and y2

are equal and coefficient of xy is zero.

The equation (3) represents a system ofcircles.

If either 1 0 or 2 0 , the equation (3)represents either the circle S = 0 or the circle S = 0.

1 2 0S S represents a system ofcircles of which S = 0, S' = 0 are members.

Consider the two circles of the system,

1 2 0S S and 3 4 0S S

1 2 3 40, 0 .

The equations of the two circles can be written

as 1 2

1 20S S

and

3 4

3 40S S

15



where 1 4 2 5

The radical axis of the two circles is given

by 3 41 2

1 2 3 40S SS S

3 4 1 2S S

1 2 3 4 0S S

1 4 2 3 0S S

0S S (independent of 1 2, ) is theradical axis of S =0 & S' = 0.

Hence every pair of circles of the system

1 2 0S S , 1 2, being parameters, has thesame radical axis.

1 2 0 S S represents a coaxal systemof circles having the radical axis S–S' = 0.

Note 1: 1 2 0S S is the equation to the coaxalsystem determined by two circles S = 0 and S' = 0.

Note 2: If the given two circles intersect, the radical axis isthe common chord and 1 2 0S S representsa system of circles passing through the points ofintersection of the circles S = 0 and S' = 0.

Note 3: If 1 2 0 i.e., 1 2 , the circles reduceto S – S' = 0 which is the radical axis of S = 0 andS' = 0.

THEOREM 2.16

If S = 0 is a circle and L = 0 is a line, then0S L ( is a parameter) represents a coaxal

system of circles of which the circle S = 0 is amember and L = 0 is the radical axis of the system.

Proof :

Let the equation of the circle be2 2 2 2 0S x y gx fy c and the line

be 0L x my n .2 2 2 2S L x y gx fy c

0x my n

2 2 2 2x y g x f m y

0c n

In this equation, the coefficients of x2 and y2

are equal and there is no xy-term. If 0 , thenS = 0 is a circle of the system. Hence, for differentvalues of , this equation represents a system ofcircles.

Now, let 1 0S L and 2 0S L be twocircles corresponding to two values of .

The radical axis is given by

1 2 0S L S L

1 2 0L

0L 1 2

This proves that every pair of circles of thesystem 0S L has the same radical axis L = 0.

0S L represents a system of coaxalcircles having the same radical axis L = 0.

Note : If the line L = 0 intersects the circle S = 0, then theequation 0S L represents the coaxal systemof circles passing through the points of intersectionof the line and the circle.

2.6 SIMPLEST FORM OF A COAXALSYSTEM :

Definition :

A coaxal system of circles is said to be inthe simplest form if its line of centres and radicalaxis are the cooridnate axes.

Note : In the simplest form of a coaxal system, usually wetake x-axis as line of centres and y-axis as radical axis.

THEOREM 2.17

The equation to a system of coaxal circlesin its simplest form is 2 2 2 0x y x c ,



is a parameter and c is a constant for all circlesof this system. (March 2005)

Proof :

Let 2 2 2 2 0x y gx fy c represents themembers of a coaxal system for different values ofg, f and c.

For the coaxal system, let the line of centresbe the x-axis and the radical axis be the y-axis.

Since the centres of all the circles of the coaxalsystem lie on the y-axis, 0 0f f .

Let the equations of any two circles of thesystem be

2 21 12 0x y g x c -- (1)

and 2 22 22 0x y g x c -- (2)

The radical axis of (1) & (2) is 1 2 1 22 0g g x c c -- (3)

Since the radical axis is the y-axis, i.e., x = 0,

1 2 1 20c c c c

Let c1 = c2 = c

The circles (1) and (2) become2 2

12 0x y g x c -- (4)

and 2 222 0x y g x c -- (5)

Any other circle coaxal with (4) & (5) willbe 2 2

32 0x y g x c

The equation of the coaxal system of circlesin the simplest form is 2 2 2 0x y x c where is a parameter and c is a constant.

Important Points :Every circle of coaxal system

2 2 2 0x y x c intersects the radical axisx = 0 at the points where y2 + c = 0, i.e., y c .Case (i) :

If c < 0, every circle of the coaxal systemintersects the radical axis x = 0 at the points 0,P c and 0,Q c . Thus every circle of

the system cuts the radical axis on two fixed pointsP and Q as in fig.

O x

y

P

Q

2.13Fig

In this case, the system is an intersectingsystem of coaxal circles.Case (ii) :

If c = 0, the two points of intersection P andQ coincide with the origin and all the circles of thesystem touch the y-axis at the origin a shown in fig.In this case, the circles form a touching system ofcoaxal circles.

x

y

O

2.14Fig

Case (iii) :If c > 0, no circle of the coaxal system

intersects the radical axis x = 0 and any two circlesof the system do not intersect at all as shown in fig.In this case, the circles form a non intersectingsystem of coaxal circles.

Ox

y

1L2L

2.15Fig

2.7 LIMITING POINTS OF A COAXAL SYSTEM

Definition :The point circles of a coaxal system are

called the limiting points of that system.

THEOREM 2.18

The limiting points of the coaxal system

17



2 2 2 0x y x c are , 0c .

Proof :Let a circle of the system be

2 2 2 0x y x c -- (1)

Centre ,0C and radius = 2 c .

For limiting points, radius = 02 20c c c

The limiting points are ,0c

Thus ,0L c and ,0L c are the two

limiting points of the coaxal system (1)Important Points :Case (i):

If c > 0, the limiting points are real and distinct.The system of circles form a non - intersecting coaxalsystem.

So for a non - intersecting coaxal system, twolimiting points exist.

The circles of the system (1) are real if radius0 2 0c

cannot lie between c and c

No circle of the system can have its centrebetween ,0L c and ,0L c . Hence thename limiting points (see fig 2.15).Note : The two limiting points are situated on the line of

centres equidistant from the radical axis oneitherside i.e., one limiting point is the image ofthe other w.r.t the radical axis of the system.

Case (ii):If c = 0 there will be only one limiting point

then all the circles of the coaxal system touch at(0, 0) (see fig 2.14).Case (iii) :

If c < 0, then the limiting points are not real.Thus the coaxal system has no limiting points.Hence an intersecting coaxal system has no limiting

points (see fig 2.13).

THEOREM 2.19Limiting points are inverse points w.r.t

every circle of the coaxal system.Proof:

Let the equation of a circle of a coaxal system

be 2 2 2 0x y x c -- (1)

Centre of the circle, ,0C ,

radius 2r c

,0L c and ,0L c are the limiting

points. Now, .CL CL c c

2 2c r

Also C, L, L' are collinear and L, L' lie on thesame side of C.

,L L are the inverse points w.r.t every circleof the coaxal system (1).Note 1: The polar of one limiting point w.r.t every circle of

the system passes through the other.Note 2: The limiting points are the conjugate points w.r.t

every circle of the coaxal system.

SOLVED EXAMPLES

1. If (–2, –1) (0, 3) are the limiting points of a coaxalsystem, find the radical axis.

Sol: The equations of the circles corresponding to thelimiting points (–2, –1) & (0, 3) are

2 22 1 0x y 2 2 4 2 5 0x y x y (S = 0)

2 20 3x y = 0

2 2 6 9 0x y y (S' = 0)

The radical axis of (1) & (2) is S – S' = 0

4 8 4 0x y 2 1 0x y 2. Find the other limiting point of the coaxial system

of which the radical axis is x = 0 and one limitingpoint is (3,-2).

Sol. Given R.A is x = 0 (y-axis) .... (1)



One limiting point is P = (3, –2)

Other limiting point ‘Q’ is image of ‘P’ w.r.t (1)

Q = (–3, –2)

3. (3, 1) is one limiting point of a coaxal system ofwhich the radical axis is 2 0x y . Find theother limiting point.

Sol: The second limiting point (h, k) is the image of(x, y) = (3, 1) w.r.t the radical aixs x – y + 2 = 0

2 3 1 23 11 1 2

h k

3 1 41 1

h k

1; 5h k

2nd limiting point (–1, 5)

4. The origin is a limiting point of a coaxal system ofwhich x2 + y2 + 2gx + 2fy + c = 0 is a member prove

that the other limiting point is 2 2 2 2,gc fc

g f g f .

(June 2003)

Sol: The equation of the circle corresponding to thelimiting point (0, 0) is x2 + y2 = 0 -- (1)

Given circle is 2 2 2 2 0x y gx fy c -- (2)

Radical axis of (1) & (2) is

2 2 0L gx fy c -- (3)

The 2nd limiting point is the image of (0, 0) w.r.t (3)

2 2

0 00 02 2 4 4

ch kg f g f

2 2 2 2, ,gc fch kg f g f

5. Find the limiting points of the coaxal systemdetermined by the circles 2 2 2 4 7 0x y x y and 2 2 4 2 5 0x y x y


2 2 2 4 7 0S x y x y -- (1)2 2 4 2 5 0S x y x y -- (2)

Radical axis of (1) & (2) is S' – S = 0

2 2 2 0x y 1 0L x y -- (3)

The equation of the coaxal system is 0S L

2 2 2 4 7 1 0x y x y x y

2 2 2 4 7 0x y x y -- (4)

Centre of (4) is 2 4,

2 2C

-- (5)

Radius of (4) is 2 22 4

74 4

r

For the limiting points, 0r 2 24 4 8 16 4 28 0

22 8 2

Substituting the values of in (5), the limitingpoints are (–2, –1) and (0, –3)

6. Find the equation of the circle which belongs to thecoaxal system determined by the limiting points(0, –3) and (–2, –1) and which is orthogonal to

2 2 2 6 1 0x y x y .

Sol: Given circle is 2 2 2 6 1 0x y x y --(1)

The limiting points are (0, –3) and (–2, –1)

The equation of the circles corresponding to the

limiting points are 2 20 3 0x y 2 2 6 9 0x y y -- (2)

and 2 22 1 0x y 2 2 4 2 5 0x y x y -- (3)

Radical axis of (2) & (3) is 4 4 4 0x y

1 0x y -- (4)The equation of any circle coaxal with (2) & (3) is

2 2 6 9 1 0x y y x y

2 2 6 9 0x y x y -- (5)

(5) is orthogonal to (1)

62 1 2 3 1 92 2

8

Equation of the required circle is2 2 8 2 1 0x y x y

7. Find the equation of the circle belonging to the coaxalsystem of which the limiting points are (1, 2), (3, 5)and which passes through (2, 2).

Sol: The equations of the circles corresponding to the

19



orthogonally.Sol: i) Let the equation of a coaxal system be

2 2 2 0 S x y x c

The limiting points of the coaxal system are ( ,0); ( ,0)A c B c

Polar of ( ,0)A c with respect to S = 0 is

( ) 0cx x c c

Now ( ) ( ) 0 c c c c c

( ,0)B c lies on ( ) 0cx x c c

Polar of A with respect to

x2 + y2 + 2 x + c = 0 passes through B

A, B are conjugate points w.r.t S = 0

ii) Equation of the circle with AB as a diameter is

x2 + y2 – c = 0

Equation to AB is y = 0

Equation of a circle passing through A , B isx2 + y2 – c + (y) = 0 (1)

Comparing x2 + y2 + 2 x + c = 0 withx2 + y2 + 2gx + 2fy + c = 0 we get

g = , f = 0, c = c

Comparing x2 + y2 + y –c = 0 withx2 + y2 + 2g1x + 2f1y + c1 = 0 we get g1 = 0,

f1 = / 2 , c1 = –c

Now 2gg1 + 2ff1 = 2( ) (0) + 2(0) (– / 2 ) = 0 = c + c1

The circle x2 + y2 + y – c = 0 is orthogonal tox2 + y2 + 2x + c = 0.

10. Let 2 2 2 0ix y g x c 1, 2, 3i be threecircles belonging to a coaxal system of circles whoseradii are r1, r2, r3 respectively. Let t1, t2 and t3 belengths of tangents to these circles from the externalpoint. Prove that

(i) 2 2 22 3 1 3 1 2 1 2 3 0g g t g g t g g t

(ii) 2 2 22 3 1 3 1 2 1 2 3g g r g g r g g r

2 3 3 1 1 2 0g g g g g g

Sol: Let (x1, y1) be an external point to the gien threecircles.

limiting points (1, 2) and (3, 5) are

2 21 2 0x y 2 2 2 4 5 0x y x y -- (1)

and 2 23 5 0x y 2 2 6 10 34 0x y x y -- (2)

Radical axis of (1) & (2) is

4 6 29 0L x y -- (3)

The equation of the circle coaxal with (1) & (2) is

2 2 2 4 5 4 6 29 0x y x y x y -- (4)

If this passes through (2, 2)

4 4 4 8 5 8 12 29 0

9 1 0 19


2 2 12 4 5

9x y x y 4 6 29 0x y

2 29 14 30 16 0x y x y

8. If A and B are the limiting points of a system of coaxalcircles, prove that the polar of either of these pointswith respect to any circle of the system passes throughthe other limiting point.

Sol: Let the equation of a coaxal system be

x2 + y2 + 2 x + c = 0The limiting points of this system are

( ,0); ( ,0)A c B c

Polar of ( ,0)A c with respect to

x2 + y2 + 2 x + c = 0 is

( ) (0) ( ) 0c y x c c

( ) 0 L cx x c c

( ,0) ( ) ( ) 0L c c c c c c

( ,0) B c lies on ( ) 0 cx x c c

polar of A with resprct to

2 2 2 0 x y x c passes through B

9. Prove that (i) the limiting points of a coaxalsystem are conjugate w.r.t every circle of thesystem.(ii) every circle through the limiting pointscuts every member of a coaxal system



Then 2 2 21 1 1 1 12t x y g x c

2 2 22 1 1 2 12t x y g x c

2 2 23 1 1 3 12t x y g x c

and 2 2 2 2 2 21 1 2 2 3 3, ,r g c r g c r g c

(i) 2 2 22 3 1 3 1 2 1 2 3g g t g g t g g t

2 22 3 1 1 12g g x y g x c

2 23 1 1 1 22g g x y g x c

2 21 2 1 1 32g g x y g x c

2 21 1 2 3 3 1 1 2x y c g g g g g g

1 1 2 3 2 3 1 3 1 22x g g g g g g g g g

2 21 1 10 2 0 0x y c x

(ii) 2 2 22 3 1 3 1 2 1 2 3g g r g g r g g r

2 22 3 1 3 1 2g g g c g g g c

21 2 3g g g c

2 3 3 1 1 2c g g g g g g

2 2 21 2 3 2 3 1 3 1 2g g g g g g g g g

2 2 2 2 2 21 2 1 3 3 2 1 2 1 3 2 30c g g g g g g g g g g g g

2 2 21 2 3 1 2 3 2 3 2 30 g g g g g g g g g g

22 3 1 1 2 3 2 3g g g g g g g g

2 3 1 2 1 3g g g g g g

2 3 1 2 3 1g g g g g g

2 2 22 3 1 3 1 1 1 2 3g g r g g r g g r

2 3 3 1 1 2 0g g g g g g

11. Show that there are only two points each of which hasthe same polar with respect to evey circle of a coaxalsystem and these are the limiting points of the system.

Sol: Let the equation of the coaxal system (in simplestform)

be 2 2 2 0x y x c , is a parameter -- (1)

Let (x1, y1) be a point such that the polar of thepoint with respect to every circle of (1) is the sameThe polar of (x1, y1) w.r.t (1) is S1 = 0

1 1 1 0xx yy x x c

1 1 1 0x x yy x c

11

1 10

x cyx y

x x

-- (2)

This represent the same line for all values of if

1 0y and 1

1

x ck

x

- (3)

(3) 1 1x c k x k

1 1 0x k c x k

The above relation is true for all values of

1 0x k and 1 0c x k

1x k and 1c x k21 1x c x c

1 1, ,0x y c

,0c are the only two points that have the samepolars w.r.t every circle of the coaxal system. Clearlythese are the limiting points of the coaxal system ofcircles (1).

EXERCISE – 2.3VERY SHORT ANSWER QUESTIONS

1. Find the radical axis of the system which hasthe limiting pointsi) (1, 4), (2, –1)ii) (1, 2), (3, 1)iii) (2, 1), (–5, –6)

2. State the nature of the coaxial system of circles 2 2 2 5 0x y x .

3. Find the equation of the system of circlescoaxal with the circles x2 + y2 = 4 and

2 2 2 4 6 0x y x y 4. Find the equation of the coaxial system

having the limiting pointsi) (3, 0), (–3,0)

ii) (0,0), (2,3)

5. i) Find the equation of the circle coaxal withthe circles 2 2 7 12 0x y x ,

2 2 8 12 0x y x which passes through(–2, 3).ii) Find the equation of the circle coaxial withthe circles 2 2 2 2 1 0x y x y and

2 2 8 6 0x y x y which passes throughthe point (–1, –2).

21



SHORT ANSWER QUESTIONS1. Find the other limiting point of the coaxal

system of which one limiting point is

i) (–2, –1) and 1 0x y is the radical axis

ii) (2, 1) and 4 0x y is the radical axis

2. Find the other limiting point of the coaxalsystem of which one limiting point is

i) (0, –3) and 2 2 2 4 7 0x y x y is amember

ii) (0, 0) and 2 2 4 4 4 0x y x y is amember (March 2006)

iii) (3, 5) and 2 2 2 2 24 0x y x y is amemberLONG ANSWER QUESTIONS

1. Prove that 2 23( ) 16 14 39x y x y 2 2 5 5 13 0x y x y represents a

coaxal system. Find the radical axis and thelimiting points of the system.

2. Find the limiting points of the coaxial system 2 2 2 13 4 0x y x x y .

3. Find the limiting points of the coaxal systemdetermined by the two circles.i) 2 2 6 8 5 0x y x y ,

2 2 8 10 5 0 x y x yii) 2 2 5 4 0x y x y

2 2 10 4 1 0x y x y (June 2002, March 2009, March 2010)

iii) 2 2 2 6 0x y x y 2 22 2 10 5 0x y y

(March 2004, May 2007, May 2009)

iv) 2 2 4 2 5 0x y x y 2 2 2 4 7 0x y x y

4. i) Find the equation of a circle which passesthrough the origin and belongs to the coaxalsystem of which (1, 2) (4, 3) are the limitingpoints. (March 2006, March 2008)

ii) The limiting points of a coaxal system ofcircles are (1, –1) and (2, 0). Find theequation of the circle of the system whichpasses through the origin. (March 2003)

iii) Find the equation of the circle belongingto the coaxial system of which the limitingpoints are (2,-3), (0,-4) and passing throughthe point (2,-1).

ANSWERS

VERY SHORT ANSWER QUESTIONS

1. i) 5 6 0x y ii) 4 2 5 0x y

iii) x + y + 4 = 02. Intersecting

3. 2 2 4 2 1 0x y x y

4. i) (x2 + y2 + 9) + x = 0

ii) x2 + y2 + (4x + 6y – 13) = 0

5. i) 2 22 25 24 0x y x

ii) 2 25 50 42 9 0x y x y


1. i) (0, –3) ii) (–5, –6)

2. i) (–2, –1) ii) (–1, –1) iii) (1, 2)


1. x–y = 0, (2, 3) (3, 2)

2. (–5, –6), (2, 1)

3. i) (1, 2), (–2, –1) ii) (–2, –1), (0, –3)

iii) (1, 2), (3, 1) iv) (–2, –1), (–7, 4)

4. i) 2 22 7 0x y x y

ii) 2 2 4 0x y y iii) 9(x2 + y2) – 52x + 46y + 105 = 0

2.8 ORTHOGONAL COAXAL SYSTEM

If two systems of coaxal circles are such thatevery member of one system cuts every member ofthe other system orthogonally, then each system issaid to be orthogonal to the other system. Two suchsystems are called conjugate coaxal systems.



THEOREM 2.20

The equation of the coaxal system ofcircles orthogonal to the coaxal system

2 2 2 0x y x c is 2 2 2 0x y y c where is a parameter..Proof :

Let 2 212 0x y x c -- (1)

and 2 222 0x y x c -- (2)

be two different circles in the given coaxalsystem of circles.

Let the circle orthogonal to (1) & (2) be

2 2 2 2 0x y gx fy k -- (3)

Where g, f, k are parameters.

(1) & (3) are orthogonal 12g c k(2) & (3) are orthogonal 22g c k

1 2 1 22 2 2 0g g g

Since (1) & (2) are different circles, 1 2

0g

0c k k c

(3) can be reduced to the form 2 2 2 0x y fy c , where f is a parameter

Thus the equation to the orthogonal coaxalsystem is 2 2 2 0x y y c , where is aparameter.Note 1 : Consider 2 2 2 0x y x c -- (A)

2 2 2 0x y y c -- (B)

The line of centres of (B) is x = 0, which is theradical axis of (A). The radical axis of (B) is y = 0,which is the line of centres of (A).(A) & (B) are called conjugate coaxal systems.

Note 2: If (A) is a non - intersecting coaxal system, then c> 0 and hence –c < 0. Thus (B) is an intersectingcoaxal system and vice versa.

Note 3: If (A) is a touching coaxal system, then c = 0 andhence (B) is also a touching coaxal system andboth the systems have the same point of contact.

Note 4: The centre and radius of (B) are 0, and 2 c . The length of the tangent from 0,

to any circle of (A). 2 20 0 c cHence the radius of any circle of one coaxal systemis equal to the length of the tangent from its centreto the circle of other coaxal system.

Coaxal System 2 2 2 0x y x c

Conjugate coaxal system 2 2 2 0x y y c

Line of centres X – axis 0y Y- axis 0x Radical axis Y- axis 0x X – axis 0y Limiting points ,0c 0, c

Points of intersections with radical axis

0, c , 0c

0c Non intersecting type 2 limiting points exist

Intersecting type Limiting points donot exist

0c Touching type of coaxal system Limiting points coincide Only one limiting point

Touching type of coaxal system Limiting points coincide Only one limiting point

0c Intersecting type Limiting points donot exist

Non – intersecting type 2 limiting points exist

23



Note : The limiting points of the coaxal system 2 2 2 0x y y c ( is a parameter) are

(0, )c . These points donot exist for c > 0.y

x

2.16Fig

y

x

2.17Fig

y

x

2.18Fig

LL

THEOREM 2.21Every circle passing through the limiting

points of a coaxal system is orthogonal to everycircle of the system.Proof :

Let the equation to the coaxal system in itssimplest form be

2 2 2 0x y x c -- (1)

(c > 0)

The limiting points of (1) are ,0L c and

,0L c .

Equation to the circle on LL' as diameter is

2 0x c x c y

2 2 0x y c

Equation of LL' is y = 0

Equation to the circle passing through L

and L' can be 2 2 0x y c ky

2 2 0x y ky c -- (2)

The condition for orthogonality of (1) & (2) is

2 0 0 0 0

2k c c

Every circle passing through the limitingpoints of a coaxal system is orthogonal to every circleof the same system.

SOLVED EXAMPLES

1. Find the equation to the system of circles orthogonalto the coxal system

2 2 3 4 2 7 0x y x y x y

Sol: Equation to given coaxal system is

2 2 3 4 2 7 0x y x y x y

2 2 3 4 2 7 0x y x y -- (1)

Let the equation of the circle orthogonal to the givencoaxal system be

2 2 2 2 0x y gx fy c -- (2)

(1) & (2) are orthogonal

3 42 2

2 2g f 2 7c

7 3 4 2 0g f g f c for all

7 0g f , 3 4 2 0g f c

Solving

12 28 21 2 4 3

g fc c

26 & 19g c f c

(2) 2 2 2 26 2 19 0x y x c y c c



3. Show that as varies, the circle

2 2 2 2x y ax by 2 0ax by form acoaxal system. Find the radical axis and also theequation of circles which are orthogonal to allcircles of the given system.


2 2 2 2 2 0x y ax by ax by -- (1)Two members of this family are

2 212 2 2 0x y ax by ax by -- (2)

and 2 222 2 2 0x y ax by ax by -- (3)

(2) - (3) 1 22 0ax by0ax by

This is independent of and it is same for every pairof circles given by (1)

(1) represents a system of coaxal circles whoseradical axis is ax – by = 0

(1) 2 2 2 1 2 1 0x y a x b y -- (4)

Let 2 2 2 2 0x y gx fy c cut (4)orthogonally

2 1 2 1 0g a b c

2 2ag bf ag bf cIf this is true for all values of , then we have

0,2c

ag bf ag bf

Solving ,4 4c c

g fa b

The system of circles orthogonal to (1) is

2 2 0

2 2c cx y x y ca b

2 22 2 0ab x y c bx ay ab

2 2 2 02c

ab x y bx ay ab

2 2 2 0ab x y bx ay ab

4. Origin is a limiting point of a system of coaxalcircles of which 2 2 2 2 0x y gx fy c is amember. Show that the equations of thecircles of the orthogonal system are 2 2 0x y g f c x y for differentvalue of y .

2 2 52 38 2 2 1 0x y x y c x yWhich is the required orthogonal system, where c is aparameter.

2. Find the equation of the orthogonal system to the

coaxal system of which 2 2 2 4 7 0x y x y is a member and (–2, –1) is a limiting point.

(July 2001)

Sol: Given circle is 2 2 2 4 7 0x y x y -- (1)

Equation of the circle correspoinding to the limiting

point (–2, –1) is 2 22 1 0x y

2 2 4 2 5 0x y x y -- (2)

Radical axis of (1) & (2) is 2 2 2 0L x y

1 0L x y

Equation of the coaxal system represented by (1)& (2) is

2 2 2 4 7 1 0x y x y x y

2 2 2 4 7 0x y x y -- (3)

Let the equation of the circle orthogonal to (3) be

2 2 2 2 0x y gx fy c -- (4)

2 4 7g f c

1 2 4 7 0g f g f c for all

1 0, 2 4 7 0g f g f c

Solving,

17 4 2 7 4 2g f

c c

3 9&

6 6c c

g f

(4)

2 2 3 9 03 3

c cx y x y c

2 23 3 3 9 3 0x y x y c x y

2 23 3 3 0x y x y c x y , which is theequation of the orthogonal system, where c is aparameter.

25



Sol: A member of the coaxal system is

2 2 2 2 0S x y gx fy c -- (1)

The equation circle corresponding to the limitingpoint (0, 0) is

2 2 0S x y -- (2)

Radical axis of (1) & (2) is L = S–S' = 0

2 2 0L gx fy c

Any circle of the coaxal system is S' + KL = 0

2 2 2 2 0x y K gx fy c -- (3)

Let the equation of the circle cutting orthogonally(1) be

2 2 2 2 0x y ax by d -- (4)

2 2a gk b fk ck d

Since (4) passes through the limiting point (0, 0) a = 0 2 2agk cbk ck

2 2

2b cag bf c g fa a

Let ba

2 2 ca g f c ag f

2 2 cb ag f

The equation of the orthogonal system is

2 2 0c cx y x yg f g f

2 2 0g f x y c x y

5. If a circle cuts orthogonally these circles S = 0,S' = 0, S" = 0, prove that it cuts orthogonally anycircle 1 2 3 0S S S

Sol: Let 2 212 0S x y g x k

2 222 0S x y g x k and

2 232 0S x y g x k

Let 2 2 2 2 0x y gx fy c cuts theorthogonally

12gg c k , 22gg c k & 32gg c k -- (3)

1 2 3 0S S S

2 2 2 21 1 2 22 2x y g x k x y g x k

2 23 32 0x y g x k

2 21 2 3 1 1 2 2 3 32x y g g g x

1 2 3 0k

2 2x y

1 1 2 2 3 3

1 2 3

2 0g g g x k

Now,

1 2 2 3 3

1 2 32 2 2 2 0

g g ggg ff g f

1 1 2 2 3 31 2 3

1 2 2 2gg gg gg

1 2 31 2 3

1 c k c k

2 2 2 2 0x y gx fy c is orthogonal to

1 2 3 0S S S

6. Show that the system of circles given by2 2 2 2 0x y x y has a common

orthogonal circle, if , , satisfy an equation ofthe form 0A B C D

Sol: Given system of circles is

2 2 2 2 0x y x y -- (1)Let the circle orthogonal to the given system be

2 2 2 2 0x y gx fy c -- (2)

2 2g f c

2 2 1 1 0g f c

0A B C D

where 2 , 2 , 1,A g B f C D c

7. A circle cuts orthogonally two fixed non-intersecting circles. Prove that it passes throughtwo fixed points on their line of centres.



equation of the coaxal system of circlesorthogonal to the given system of coaxalcircles.

4. Show that the equation 2 2 2 1x y x 2 1 2 2 0y where is a parameter

represents a coaxal system. Find the equationof a member of this system, which cutsorthogonally the circle x2+ y2 + 4x–6y + 8 = 0.

ANSWERS


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

ii) 2 25 14 6 8 2 5 0x y x y x y

2. 2 23 4 0, 4 3 24 0x y x y x y

3. i) 2 21 4 0x y x y

ii) 2 2 2 3 3 0x y y x y

4. 2 2 9 0x y x

ADDITIONAL INFORMATION

Result 1 :

If is the angle between the two intersectingcircles whose radii are r1, r2 then length of

the common chord is

1 2

2 21 2 1 2

2 sin

2 cos

r r

r r r r

Explanation :

Let PQ = 2h

If ' ' is angle between two circles then anglebetween their radii is 180 – .

C1 C2

P

Q

hh

Area of the triangle PC1C2

Sol: Let the two non - intersecting circles be

2 212 0S x y x c and

2 222 0S x y x c

12g c k ....... (1) and 22g c k -- (2)

(1) - (2) 1 22 0g 0g 1 2

k c The equation of the circle cutting 0 & 0S S

orthogonally 2 2 2 0x y fy cThe line of centres of S = 0 & S' = 0 is y = 0Since 0 & 0S S are non - intersecting, c > 0

0 ,0c c are real

The circle 2 2 2 0x y fy c passes through twofixed points ,0c on the line of centres.

EXERCISE – 2.4LONG ANSWER QUESTIONS

1. i) Find the equation to the system ofcircles orthogonal to the coaxal system

2 2 3 4 2 1 0x y x y x y , is a parameter.

ii) Find the coaxal system which is orthogonalto the coaxal system 2 2 2 3 1x y x y 4 1 0x y , is a parameter.. (June 2005)

2. Show that the circles given by 2 2 6 8x y x y 3 4 0x y , ( is a

parameter) forms a coaxal system and find theequation of common radical axis. Find, alsothe coaxal system orthogonal to the givencoaxal system.

3. i) The origin is a limiting point of a system ofcoaxal circles of which x2 + y2 + 2x + 2y – 4 = 0is a member. Find the equation of the coaxalsystem orthogonal to the given coaxal system.

ii) The point (–2, –1) is the limiting point of acoaxal system of which

2 2 8 2 1 0x y x y is member. Find the

27



= 1 2 1 21 1( ) sin(180 )2 2

C C h r r

1 2

2 21 2 1 2

sin

2 cos

r rhr r r r

Length of the common chord = 2h

=

1 2

2 21 2 1 2

2 sin

2 cos

r r

r r r r

Note 1 : If is angle between the radii of two circles then

length of the common chord is

1 2

2 21 2 1 2

2 sin

2 cos

r r

r r r r

Note 2 : The length of the common chord of two circles of

radii r1 and r2 which intersect at right angles is 1 2

2 21 2

2r r

r r

Result 2 :

If A and B are conjugate points w.r.t. thecircle S = 0 then the circle having AB asdiameter cuts the circle S = 0 orthogonally.

Explanation :

Let A(x1, y1), B(x2, y2)

Since, A, B are conjugate points w.r.t the circleS = 0 then S12 = 0

i.e., x1x2 + y1y2 + g(x1 + x2) + f(y1 + y2) + c = 0

Equation of circle having AB as diameter is

x2 + y2 – (x1 + x2)x – (y1 + y2)y + x1x2 + y1y2 = 0

For the above circle and S = 0

2gg' + 2ff ' = –g(x1 + x2) – f(y1 + y2)

= x1x2 + y1y2 + c = c'+ c

Hence proved.Note : Converse of the above theorem is also true.

Ex : If the points A(2, 3) and B(–7, –12) are conjugate pointsw.r.t. the circle x2 + y2 – 6x – 8y – 1 = 0. Then find the centreof the circle passing through A and B orthogonal to givencircle.

Sol. From the above theorem required centre is mid point

of AB i.e.,

5 9,

2 2

Result 3 :

Distance from one of the limiting points ofthe coaxial system 0S L to its radical axisis 2 2d r . where d is the perpendiculardistance from centre of S = 0 to the radical axisand 'r' is radius.

Explanation :

Since from any point on the radical axis, lengthsof tangents to all the circles are equal and limitingpoint is also one of the circle of co-axial system.

r

dL.P.

R.A.

C

Distance of R.A. from limiting point isequal to length of the tangent to the circle from apoint on the R.A.

i.e., 2 2d r

Note : Distance between two limiting points of coaxial

system is 2 22 d r .

Result 4 :

In the co-axial system of circles, commontangent subtend right angle at their limitingpoint (if exists) of co-axial system.

Explanation :Radical axis bisects common tangents,The circle whose common tangent as a diameter



is a member of orthogonal system of givensystem of circles ( centre of the circle lies onthe radical axis of given coaxial system)

900

C1 C2L1 L2

That circle passing through limiting pointsof coaxial system.

( every circle of orthogonal system passingthrough limiting points of given system ofcircle)

Common tangents subtends right angle atthe limiting points.

Result 5 :

Each side of ABC is the polar of theopposite vertex w.r.t a circle with centre P.For ABC , the point P is the orthocentre.

Explanation :

P

A

C

B

Polar of A is BC which is perpendicular to PA

Polar of B is CA which is perpendicular to PB

Polar of C is AB which is perpendicular to PC

P is the orthocentre of ABC

Result 6 :

The radical centre of three circles describedon the three sides of a triangle as diameteris its orthocentre.

Explanation :A

CB

C1 C3

C2

Let C1, C2, C3 are the midpoints of AB, BC, CArespectively.

Let P be the radical centre then PA C1C3 || BC( Line joining the midpoints of the sides of thetriangle is parallel to third side)

PA BC

Similarly PB CA & PC AB

P is the orthocentre of ABCResult 7 :

A, B, C are the centres of three circles of equal radiiwhich do not touch externally pairwise and whosecentres are non-collinear. Then the radical centre ofthe circle is circumcentre of ABC .

Explanation :

Method - I :

Let P be the radical centre of three circles.

S11 = S'11 = S"11

AP2 – r12 = BP2 – r2

2 = CP2 – r32

r1 = r2 = r3

AP = BP = CP

P is the circumcentre of ABC

Method - II :

P

A B

C

C1 C2

C3

29



Since radii of C1 & C2 are equal, the radicalaxis of C1 & C2 is perpendicular bisector ofAB

Similarly remaining also.

P is the circumcentre of ABC

Result 8 :

If A, B, C are the centres of three circlestouching mutually extrenally. Then theradical centre of the circles is incentre ofABC .

Explanation :

A C

B

PP1 P2

P3

Let P be the radical centre.

Let P1, P2, P3 be the point of contacts.then 1 2 3, ,PP AB PP BC PP CA and

1 2 3PP PP PP (length of tangents) P is incentre of ABC

Result 9 :

A, B, C are the centres of three circles C1,C2, C3 such that C1, C2 touch each otherexternally and they both touch C3 frominside. Then the radical centre of the circlesis Excentre opp to C of ABC

Explanation : 1 2 3, ,PP AB PP CA PP CB

& 1 2 3PP PP PP

P is excentre opp to C of ABC

A B

C

P

P3P2 P1

Note 1 : C2, C3 touch each other externally and both touchC1 internally then radical centre is excentre oppositeto A of ABC

Note 2 : C3, C1 touch each other externally and both touchC2 internally, then radical centre is Excentre oppositeto B of ABC .

ADDITIONAL SOLVED EXAMPLES

1. If 2 4 0,S x y x 2 8 0S x y x andAB is a direct common tangent to the two circleswhere A,B are points of contact then find the anglesubtended by AB at origin.

Sol: Clearly the given circles are touch each other exter-nally and origin is the point of contact of two circles.

Which is the limiting point of the coaxl system of

which S = 0 & S'= 0 are members.A B

P

Since direct common tangent subtends a right

angle at limiting point the angle subtended by AB

at origin is 2

2. Find radical centre of three circles discribed onthe three sides 4x – 7y + 10 = 0, x+y – 5 = 0 and7x + 4y – 15 = 0 of triangle as diameters

Sol: The Radical centre of three circles discribed in the

three sides of a triangle as diameters is the orthocentre

of the triangle

The orthocentre of the triangle formed by the lines

4x – 7y + 10 = 0, x+y – 5 = 0 and 7x + 4y – 15 = 0 isthe point of intersection of the 'r sides4x – 7y +10 = 0 & 7x + 4y – 15 = 0



By solving these two we get (1,2) Radical centre is (1,2)

3. In a ABC B = (2,3), C= (7,–1) and AB = 3, AC = 5then Find the equation of altitude through the ver-tex A.

Sol: Given B = (2,3), C= (7, –1) and AB = 3 , AC = 5

Equation of the circle having B as centre and AB = 3

as radius is (x-2)2 + (y–3)2 = 9

A

B C

i.e x2 + y2 – 4x – 6y + 4 = 0 ------ (1)

Equation of the circle having C as centre and CA = 5

as radius is (x – 7)2 + (y + 1)2 = 25i.e x2 + y2 – 14x + 2y + 25 = 0 ........ (2)

now altitude of the ABC through A is the radicalaxis of circles (1) & (2)Since radical axis is to line of centres Equation of the altitude is S – S' = 0

i.e 10x – 8y = 214. The circles having radii 1,2,3 touch each other ex-

ternally then find the radius of the circle which cutsthe three circles orthogonally.

Sol: Let A,B,C are the centres of three circles having radii

1,2,3 respectively

AB = 3 , BC = 5, CA = 4

AB

C

The circle cuts the given three circles orthogonallyis the incircle of ABC Required radius is the inradius of ABCperimeter = 2s = 3 + 4+ 5 = 12 6s

Area = ( )( )( ) 6s s a s b s c

1rs

*5. Find the length of the common chord of the circle

2 2 22 0x y hx a and

2 2 22 0x y ky a .Sol: S x2 + y2 + 2hx + a2 = 0 -------- (1)

'S x2 + y2 – 2ky – a2 = 0 -------- (2)Equation of common chord is S – S' = 0 hx + ky + a2 = 0centre of (1) = (–h, 0)

radius of (1), 2 2r h a

Now, 2 2 2 2

2 2 2 2

h a h adh k h k

length of common chord

2 22 r d

2 2 2 2

2 22a k h a

h k

*6. Two circles are drawn through the points (1,5) and(4,1) to touch the axis of y. Find the angle at whichthey intersect.

Sol : Suppose the equation of the required circle is

x2 + y2 + 2gx + 2fy + c = 0

y-axis is a tangent c = f 2

Equation of the circle becomes

x2 + y2 + 2gx + 2fy +f 2 = 0

This circle passes through A(1, 5) and B(4, 1)

1 + 25 + 2g + 10f + f 2 = 0 (1),

16 + 1 + 8g + 2f + f 2 = 0 (2)

4(1) – (2) 3f 2 + 38f + 87 = 0

3f 2 + 9f + 29f + 87 = 0

3f (f + 3) + 29 ( f + 3) = 0

(3f + 29)(f + 3) = 0

3f + 29 = 0 or f + 3 = 0

f = –29/3 or f = –3

Case (i) : suppose f = –3

Substituting in (1), we get 26 + 2g – 30 + 9 = 0 2g = 30 – 35 = –5 g = –5/2 and c = f 2 = 9 Equation of the required circle is

x2 + y2 – 5x – 6y + 9 = 0Case (ii) : Suppose f = –29 / 3

31



Substituting in (1), we get 290 84126 2 0

3 9 g

290 8412 263 9

g 870 841 234 205

9 9

and 2

2 29 8413 9

c f


2 2 205 58 8410

9 3 9 x y x y

1 1 1

2 2 2 2

2 2cos2 ' ' '

c c gg ff

g f c g f c

2

841 5 205 299 2 2( 3)9 2 18 3

25 205 841 8412 9 94 18 9 9

81 841 1025 589 18

5 20522 18

1844 1025 1044225 918

5 205 5 205 4118

1 9cos41

*7. Find the equation of the circle having a radius2 and passing through the limiting points of thecoaxial system 2 2 6 2x y 4 0x y .

Sol: Equation of coaxial system is

x2 + y2 – 6 – 2 (x + y – 4) = 0On computing, limiting points are (1,1), (3,3).Let C = (a,b) be the centre of the required circleNow CP = CQ

CP (1,1)

Q (3,3) a + b = 4and CP = 2

a2 + b2 = 10C = (1,3) or (3,1)Hence equations of required circles isx2 + y2 – 2x – 6y + 6 = 0

x2 + y2 – 6x – 2y + 6 = 0*8. If a circle passes through the point (a,b) and cuts

x2 + y2 = k2 orthogonally then find the equation ofthe locus of its centre.

Sol: Equation of given circle is x2 + y2 – k2 = 0 --(1)

Let the equation of required circle be

x2 + y2 + 2gx + 2fy + c = 0 --(2)(1) and (2) are perpendicular

0 + 0 = – k2 + c

2c k (2) is passing through (a,b)a2 + b2 + 2ga + 2fb + k2 = 0 locus of centre (–g, –f) is a2 + b2 – 2ax – 2by + k2 = 0 2ax + 2by – (a2 + b2 + k2) = 0

ADDITIONAL EXERCISE*1. If a circle passes through (1,2) and cuts

x2 + y2 = 4 orthogonally then show that theequation of the locus of centre is

2 4 9 0x y .

2. Prove that the two circles which pass throughthe points (0, a), (0, –a) and touch the liney = mx + c will cut orthogonally if

2 2 22c a m

3. The centre of the circle S = 0 lie on the line2 2 9 0x y and S = 0 cuts orthogonallythe circle x2 + y2 = 4. Show that the circlepasses through two fixed points (–4, 4) and

1 1,2 2

.

4. A point P moves so that the distances fromtwo fixed points are in a constant ratio 1 .Prove that the locus of P is a circle. If varies, show that P generates a system ofcoaxals of which the fixed points are thelimiting points.

5. Prove that the polars of any point with respect toa system of coaxal circles pass through a fixedpoint.



6. Show that the equation of the circle passingthrough (1, 1) and the points of intersection of

the circles 2 2 13 3 0x y x y and2 22 2 4 7 25 0x y x y is2 24 4 30 13 25 0x y x y .

7. If a circle passes through the point (a, b) andcuts the circle x2 + y2 = k2 orthogonally, showthat the locus of its centre is

2 2 22 2 0ax by a b k .

8. If the circles 2 2 2 2 6 0x y x ky and2 2 2 0x y ky k intersect orthogonally,,

show that k = 2 or –3/2.9. Show that the distance of the centre of the circle

x2 + y2 = 2x from the common chord of thecircles 2 2 5 8 1 0x y x y and

2 2 3 7 25 0x y x y is 2.

10. If the circles 2 2 2 0x y ax cy a and2 2 3 1 0x y ax dy intersect in two

distinct points P and Q then the line5 6 0x y a passes through P and Q forno value of a .

11. Circles of radii 3,4,5 touch each otherexternally. Let P be the point of intersectionof tangents to these circles at their points ofcontact. Show that the distance of P from apoint of contact is 5 .

12. Three circles touch each other externally. Thetangents at their points of contact meet at a pointwhose distance from a point of contact is 4.Show that the ratio of the product of the radiito the sum of the radii of the circles is 16.

13. Show that the equation of the line passingthrough the points of intersection of the circles

2 23 3 2 12 9 0x y x y and 2 2x y x y 2 2 6 2 15 0x y x y is 10 3 18 0x y .

14. Two circles x2 + y2 = 6 and x2 + y2– 6x +8 = 0are given show that the equation of the circlethrough their points of intersection and the point

(1, 1) is 2 2 3 1 0x y x .

15. Consider a family of circles passing throughthe points (3, 7) and (6, 5). Show that thechords in which the circle

2 2 4 6 3 0x y x y cuts the members of

the family are concurrent at a point 232,3

.

16. A circle touches the line 2 3 1 0x y at thepoint (1, –1) and is orthogonal to the circlewhich has the line segment having end points(0, –1) and (–2, 3) as diameter. Show that theequation of the circle is

2 22 2 10 5 1 0x y x y .

17. A circle C of radius unity lies in the firstquadrant and touches both the coordinate axes.Show that the radius of the circle which touchesboth coordinate axes, lies in the first quadrantand cuts C orthogonally is 2 3 .

18. From a point A(3, 2), tangents are drawn to

the circle 2 2 2 4 4 0x y x y . If BC isthe chord of contact of tangents, show that theline joining the mid - points of AB and AC is4 8 17 0x y .

19. Show that the locus of the centres of the circlespassing through the points of intersections ofthecircles 2 2 1x y and 2 2 2 0x y x y

is 2 0x y

20. Show that the equation of the circle whichpasses through the point (–1, 7) and whichtouches the straight line x = y at (1, 1) is

2 2 3 7 2 0x y x y

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SYSTEM OF CIRCLES MATHEMATICS · Note : The angle between two circles at either point of intersection is the same. THEOREM 2.1 Let C 1, C 2 be the centres of two intersecting circles