1. Find the radical centre of the circles x2 + y2 - 4x - 6y + 5 = 0, x2 + y2 - 2x- 4y- 1 = 0; x2 + y2 - 6x- 2y = 0.
Sol: Given Circles: (S) : x2 + y2 - 4x - 6y + 5 = 0
(S') : x2 + y2 -2x-4y-1 = 0
(S'') : x2 + y2- 6x- 2y = 0
S - S' = 0 -2x -2y + 6 = 0
⇒ x + y -3 = 0 ................. (1)
S'- S'' = 0 ⇒ 4x -2y-1 = 0 ................. (2)
Solving (1) and (2) we get the radical centre.
2. If the circles x2 + y2 + 2ax + c = 0 and x2 + y2 + 2bx + c = 0 touch each other then show that 
.
Sol: Given first Circle : x2 + y2 + 2ax + c = 0
Centre : C1(-a, 0)
3. Find the equation of the circle Coaxial with the circles x2 + y2 - 2x + 2y + 1 = 0 and x2 + y2 + 8x - 6y = 0 which passes through the point (-1, -2) .
Sol: Given circles : (S): x2 + y2 -2x + 2y + 1 = 0
(S'): x2 +y2 + 8x- 6y = 0
Radical axis (L) : 
⇒ -10x + 8y + 1 = 0
⇒ 10x - 8y - 1 = 0
Any circle of coaxial system of circle : S + λL = 0
⇒ x2 + y2 - 2x + 2y + 1 + λ (10x - 8y - 1) = 0
⇒ x2 + y2 + (10λ - 2) x + (2– 8λ) y+ (1 - λ) = 0
Since this circle passes through (-1, -2)
⇒ 1 + 4 -1 (10λ - 2) -2 (2 - 8λ) + (1 - λ) = 0
⇒ 5 - 10λ + 2 - 4 + 16λ + 1 - λ = 0
⇒ 5λ + 4 = 0
Required circle : x2 + y2 - 2x + 2y + 1 - 4/5 ( 10x - 8y -1) = 0
⇒ 5(x2 + y2) - 10x +10y +5 - 40x + 32y + 4 = 0
⇒ 5(x2 + y2) - 50x + 42y + 9 = 0
4. Find the equation of the circle which is coaxial with the circles x2 + y2 -6x + 4 = 0 and x2 + y2 -5x + 4 = 0 and touches the line 3x - 4y = 15 .
Sol: Given circles:
(S) : x2 + y2 -6x + 4 = 0
(S') : x2 + y2 -5x + 4 = 0
Radical axis (L) :
⇒ -x = 0
⇒ x = 0
Any circle of coaxial system of circle : S + λL = 0
⇒ x2 + y2 -6x + 4 + λ (x) = 0
⇒ x2 + y2 + x (λ-6) + 4 = 0
Squaring on both sides
⇒ 9λ2 + 144 + 72λ = 25λ2 - 300λ + 500
⇒ 16λ2 -372λ + 356 = 0
⇒ 4λ2 - 93λ + 89 = 0
⇒ 4λ2 - 4λ - 89λ + 89 = 0
⇒ 4λ(λ -1) - 89 (λ -1) = 0
⇒ (λ -1) (4λ - 89) = 0
λ = 89/4
∴ λ = 1;
Required circles: λ = 1 ⇒ x2 + y2 -6x + 4 + 1 (x) = 0
⇒ x2 + y2 -5x + 4 = 0
λ = 89/4 ⇒ x2 + y2 -6x + 4 + (x) = 0
⇒ 4( x2 + y2) - 24x + 16 + 89x = 0
⇒ 4( x2 + y2) + 65x + 16 = 0
5. Find the angle between the circles x2 + y2 + 4x -14y + 28 = 0 and x2 + y2 + 4x - 5 = 0.
Sol: Given first circle : (S) : x2 + y2 + 4x -14y + 28 = 0 ; Centre : C1 (-2, 7)
∴ θ = 60°
6. Find the equation to the circle whose diameter is the common chord of the two circles x2 + y2 - 4x + 6y -12 = 0 and x2 + y2 + 2x - 2y - 23 = 0. Hence find the length of the common chord.
Sol: Given Circles : (S) : x2 + y2 - 4x + 6y-12 = 0 ; (S' ) : x2 + y2 + 2x- 2y- 23 = 0
Common chord (L) :
⇒ -6x + 8y + 11 = 0
⇒ 6x - 8y - 11 = 0
Any circle of coaxial system of circle : S + λL = 0
⇒ x2 + y2 - 4x + 6y-12 + λ (6x - 8y- 11) = 0
⇒ x2 + y2 + x(6λ - 4) + y (6 - 8λ) + (-12 - 11λ) = 0
Centre : (2 - 3λ , 4λ - 3)
By data : 6(2 - 3λ) -8 (4λ-3) -11 = 0
⇒ 12 -18λ - 32λ + 24 - 11 = 0
⇒ -50λ + 25 = 0
Required Circle : x2 + y2 - 4x + 6y - 12 + 1/2 (6x - 8y - 11) = 0
⇒ 2(x2 + y2) - 8x + 12y - 24 + 6x - 8y - 11 = 0
⇒ x2 + y2 - x + 2y - 35/2 = 0
2g = -1 ; 2f = 2 ; C = - 35/2
g = - 1/2 ; f = 1
7. Find the equation of the circle which is orthogonal to the circles x2 + y2 + 2x +17y + 4 = 0, x2 + y2 + 7x + 6y + 11 = 0 and x2 + y2 - x + 22y + 3 = 0.
Sol: Let the required circle be : x2 + y2 + 2gx + 2fy + c = 0 .
Given Circle : x2 + y2 + 2x + 17y + 4 = 0
g' = 1; f'' = 17/2 ; c' = 4
Orthogonal Condition : 
⇒ 2g(1) + 2f (17/2 ) = c + 4
⇒ 2g + 17 f - c = 4 ................ (1)
Given Circle : x2 + y2 + 7x + 6y + 11 = 0
g' =7/2 ; f ' = 3 ; c' = 11
Orthogonal condition : 2g ( 7/2 ) + 2f(3) = c + 11
7g + 6f - c = 11 ................ (2)
Given Circle : x2 + y2 - x + 22y + 3 = 0
g' = -1/2 ; f'' = 11; c' = 3
Orthogonal condition : 2g (-7/2) + 2f (11) = c + 3
⇒ -g + 22f - c = 3 ................ (3)
from (1) = 2(-3) + 17(-2) - c = 4
⇒ - 40 - c = 4
⇒ c = - 44
Required circle : x2 + y2 + 2(-3) x + 2(-2) y - 44 = 0
⇒ x2 + y2 - 6x - 4y - 44 = 0
8. Find the equation of the circle which passes through the point (0, -3) and intersects the circles given by the equations x2 + y2 - 6x + 3y + 5 = 0 and x2 + y2 - x- 7y = 0 orthogonally.
Sol: Let the required circle be: x2 + y2 + 2gx + 2fy + c = 0
Since this circle passes through (0, -3)
⇒ 0 + 9 + 2g(0) + 2f(-3) + c = 0
⇒ -6f + c = -9 .................. (1)
Given Circle : x2 + y2 - 6x + 3y + 5 = 0
g' = -3 ; f'' = 3/2 ; c' = 5
Orthogonal Condition :
2g(-3) + 2f (3/2 ) = c + 5
-6g + 3 f - c = 5 .................. (2)
Given circle : x2 + y2 - x - 7 y = 0
g' = - 1/2; f ' = -7/2 ; c' = 0
Orthogonal condition : 2g (-1/2 ) + 2f (- 7/2 ) = c + 0
⇒ -g -7 f- c = 0
⇒ g + 7 f + c = 0 .................. (3)
Solving (1) and (2)
9. Find the equation of the circle which pass through the origin, having its centre on the line x + y = 4 and intersecting the circle x2 + y2 - 4x + 2y + 4 = 0 orthogonally.
Sol : Let the required Circle be : x2 + y2 + 2gx + 2fy + c = 0
Since this Circle passes through (0, 0)
Centre (-g, -f) lies on the line : x + y = 4
⇒ -g - f = 4
⇒ g + f = - 4 .................. (1)
Given Circle : x2 + y2 - 4x + 2y + 4 = 0
g' = -2 ; f'' = 1 ; c' = 4
Orthogonal Condition :
2g(-2) + 2f(1) = 0 + 4
⇒ - 4g + 2f = 4
⇒ 2g - f = -2 .................. (2)
Solving (1) and (2)
g + f = - 4
2g - f = - 2
---------------------
⇒ 3g = - 6
from (1) : -2 + f = -4
Required circle : x2 + y2 + 2(-2) x + 2(-2) y + 0 = 0
⇒ x2 + y2 - 4x - 4y = 0
10. Find the limiting points of the coaxial system, determined by the circles x2 + y2 + 10x - 4y - 1 = 0 and x2 + y2 + 5x + y + 4 = 0.
Sol: Given circle (S) : x2 + y2 + 10x - 4y - 1 = 0
(S') : x2 + y2 + 5x + y + 4 = 0
Radical axis (L) : S - S' = 0
⇒ 5x - 5y - 5 = 0
⇒ x - y - 1 = 0
Any circle of coaxial system of circle : S + λ L = 0
x2 + y2 + 10x - 4y- 1+ λ (x - y - 1) = 0
x2 + y2 + (λ + 10) x + (- 4 - λ) y + (-1 - λ) = 0
⇒ λ2 + 100 + 20λ + λ2 + 16 + 8λ + 4 + 4λ = 0
⇒ 2λ2 + 32λ + 120 = 0
⇒ λ2 + 16λ + 60 = 0
⇒ λ2 + 6λ + 10λ + 60 = 0
⇒ λ(λ + 6) + 10 (λ + 6) = 0 (λ + 6) (λ + 10) = 0
⇒ λ = -6; λ = -10
Substituting the values of λ in centre, we get
Limiting points : λ = -6 ⇒ (-2, -1)
λ = -10 ⇒ (0, -3)
11. The point (2, 1) is a limiting point of a coaxial system of circle of which x2 + y2 - 6x - 4y - 3 = 0 is a member. Find the radical axis and the other limiting point.
Sol: Given limiting point : (2, 1)
Then circle (S) : (x - 2)2 + (y - 1)2 = 0
⇒ x2 + y2 - 4x - 2y + 5 = 0
Given Circle (S') : x2 + y2 - 6x - 4y - 3 = 0
Radical axis (L) ; S - S' = 0
⇒ 2x + 2y + 8 = 0
⇒ x + y + 4 = 0
Let Other limiting point be: (h, k) ; Then
⇒ h - 2 = k - 1 = -7
⇒ h - 2 = -7 ; k -1 = -7
⇒ h = -5 ; k = -6
... Other limiting point = (-5, -6)
12. The origin is a limiting point of a coaxial system of which x2 + y2 + 2gx + 2fy + c =0 is a member prove that other limiting point is 
.
Sol: Given Limiting Point: (0, 0) ; Then circle (S) : (x - 0)2 + (y - 0)2 = 0
⇒ x2 + y2 = 0
Given circle (S') : x2 + y2 + 2gx + 2fy + c = 0
Radical axis (L) : S - S' = 0
⇒ - 2gx - 2fy- c = 0
⇒ 2gx + 2fy + c = 0
Any circle of coaxial system of circle : S + λL = 0
⇒ x2 + y2 + λ(2gx + 2fy+ c) = 0
⇒ x2 + y2 + 2gλx + 2fλy + cλ = 0
Centre : (- gλ, - fλ)
Radius : 
By data : Radius = 0
⇒ g2λ2 + f2λ2 - cλ = 0
⇒ λ2 (g2 + f2) - cλ = 0
⇒ λ [ λ (g2 + f2) - c] = 0
λ = 0; λ (g2 + f2) - c = 0
13. Find the equation of the circle which passes through the origin and which belongs to the coaxial system of which the liming points are (1, 2) and (4, 3).
Sol: Given Limiting Points : (1, 2) and (4, 3)
From (1, 2) : Circle (S) : (x - 1)2 + (y - 2)2 = 0
⇒ x2 + y2 - 2x - 4y + 5 = 0 .................. (1)
From (4, 3) : (S' ) (x - 4)2 + (y - 3)2 = 0
⇒ x2 + y2 - 8x - 6y + 25 = 0 .................. ( 2)
Radical Axis (L) : S - S' = 0
⇒ 6x + 2y - 20 = 0
⇒ 3x + y - 10 = 0
Any circle of coaxial system of circle : S + λL = 0
⇒ x2 + y2 - 2x - 4y + 5 + λ ( 3x + y - 10) = 0
Since it passes through (0, 0 )
⇒ 5 - 10λ = 0
Required Circle : x2 + y2 - 2x - 4y + 5 + 
(3x + y - 10) = 0
⇒ 2 (x2 + y2) - 4x - 8y + 10 + 3x + y - 10 = 0
⇒ 2 (x2 + y2) - x - 7y = 0
14. (Find the equation of the circle which belongs to the coaxial system determined by the limiting points (0, - 3) and (- 2, - 1) and which is orthogonal to the circle x2 + y2 + 2x + 6y + 1 = 0.
Sol: Given Limiting Points : (0, -3) and (-2, -1)
From (0, - 3) : Circle : (S) : (x - 0)2 + (y + 3)2 = 0
⇒ x2 + y2 + 6y + 9 = 0 .................. (1)
From (- 2, - 1) , Circle : (S' ) : (x + 2)2 + (y + 1)2 = 0
⇒ x2 + y2 + 4x + 2y + 5 = 0 .................. ( 2)
Radical Axis (L) : S- S' = 0
⇒ - 4x + 4y + 4 = 0
x - y - 1 = 0
Coaxial System of Circle : S + λL = 0
⇒ x2 + y2 + 6y + 9 + λ (x - y- 1) = 0
⇒ x2 + y2 + λx + (6 - λ) y + (9 - λ) = 0 .................. (3)
Given Circle : x2 + y2 + 2x + 6y + 1 = 0 .................. (4)
g' = 1; f'' = 3; c' = 1
Orthogonal Condition for (3) and (4) :
⇒ λ + 18 - 3λ = 10 - λ
⇒ 18 - 2λ = 10 - λ
Required Circle : x2 + y2 + 6y + 9 + 8 (x - y - 1) = 0
⇒ x2 + y2 + 6y + 9 + 8x - 8y - 8 = 0
⇒ x2 + y2 + 8x - 2y + 1 = 0
15. Find the equation to the system of circle orthogonal to the coaxial system x2 + y2 + 3x + 4y - 2 + λ (x + y - 7) = 0.
Sol: Equation to given coaxial System is
x2 + y2 + 3x + 4y - 2 + λ (x + y - 7) = 0
⇒ x2 + y2 + x (3 + λ) + y(4 + λ) + (- 2 - 7λ) = 0 .................. (1)
Let the Equation to the Circle Orthogonal to the given Coaxial
System be : x2 + y2 + 2gx + 2fy + c = 0 .................. ( 2 )
By data: (1) and (2) are Orthogonal
⇒ g (3 + λ) + f (4 + λ) = c - 2 - 7λ
⇒ λ (g + f + 7) + 3g + 4f - c + 2 = 0
λ ⇒ g + f + 7 = 0 .................. ( 3 )
3g + 4f + 2 - c = 0 .................. ( 4 )
solving (3) and (4)
g = - 26 - c; f = 19 + c
Equation (2) becomes
x2 + y2 + 2 (- 26 - c) x + 2(19 + c) y + c = 0
⇒ x2 + y2 - 52x + 38y - c (2x - 2y - 1) = 0
Which is the required orthogonal system, when 'C' is a parameter.
16. Prove that the limiting points are inverse points with respect to every circle of coaxial system.
Sol: Let the equation of a circle of a coaxial system be : x2 + y2 + 2λx + c = 0.
Centre : C ( - λ , 0)
Radius : 
By data : Radius = 0
⇒ λ2 - c = 0
⇒ λ2 = c
L (- 
, 0) and L' ( , 0) are Limiting points
CL = - λ +
CL' = - λ -
CL . CL' = (- λ + ) (- λ - )
CL . CL' = λ2 - c
CL . CL' = r2
also C, L, L' are Collinear and L, L' lie on the same circle of C
Hence, L and L' are Inverse points w.r.t. every Circle of the system.
17. Prove that the equation of the circle which cuts orthogonally each member of a given coaxial system x2 + y2 + 2 λx + c = 0 is x2 + y2 + 2fy - c = 0.
Sol: The given coaxial system is given by the equation :
x2 + y2 + 2λx + c = 0 .................. (1)
Consider tow circle of this System for values λ1 and λ2 of λ.
x2 + y2 + 2λ1x + c = 0 .................. (2)
x2 + y2 + 2λ2x + c = 0 .................. (3)
x2 + y2 + 2gx + 2fy + k = 0
Let the Circle Orthogonal to (2) and (3) be ... 2 λ1g = c + k ; 2λ2g = c+k
⇒ 2 (λ1 - λ2) g = 0
⇒ g = 0 [... λ1 ≠ λ2]
... k = - c
Hence the equation of the circle which cuts every member of the given coaxial system orthogonally is x2 + y2 + 2fy - c = 0 .
18. Show that the equation x2 + y2 + 2 (1 - λ) x + 2 (1 - 2λ) y - 2 = 0 where λ is a parameter, represents a system of Coaxial Circles. Also find the equation of the circle which cuts orthogonally the circle x2 + y2 + 4x - 6y + 8 = 0.
Sol: Given equation : x2 + y2 + 2 (1 - λ) x + 2 (1 - 2λ) y - 2 = 0
⇒ x2 + y2 + 2x + 2y - 2 - λ(2x + 4y) = 0
This clearly represents a circle
Let λ1 and λ2 be two values of the parameter λ, then the two circles corresponding to two values of λ are x2 + y2 + 2x + 2y - 2 - λ1 ( 2x + 4y ) = 0 .................. (2)
and x2 + y2 + 2x + 2y - 2 - λ2 ( 2x + 4y) = 0 .................. (3)
Radical axis of (2) and (3) : ( λ1 - λ2 ) ( 2x + 4y ) = 0
2x + 4y = 0 [... λ1 ≠ λ2]
That is every pair of circle S + λL = 0 has the same radical axis 2x + 4y = 0
Equation (1) represents a coaxial system of circles
Given another circle : x2 + y2 + 4x - 6y + 8 = 0 .................. (4)
(1) and (4) cut each other orthogonally.
⇒ 2 (1 - λ) (2) + 2 (1 - 2 λ) ( - 3) = - 2 + 8
⇒ 2 - 2 λ - 6 + 12 λ = 6
⇒ 10 λ - 4 = 6
⇒ 10 λ = 10
⇒ λ = 1
The equation of the circle cutting (4) orthogonally and belonging to (1) is
x2 + y2 + 2x + 2y - 2 - ( 2x + 4y ) = 0
⇒ x2 + y2 - 2y- 2 = 0
19. (- 2, - 1) is a limiting point of a coaxial system of which x2 + y2 + 2x + 4y + 7 = 0 is a member find the equation of the orthogonal system.
Sol: Given Limiting point : (- 2, - 1)
From (- 2, - 1) : Circle (S) : (x + 2)2 + (y + 1)2 = 0
⇒ x2 + y2 + 4x + 2y + 5 = 0 .................. (1)
Given circle (S' ) : x2 + y2 + 2x + 4y + 7 = 0 .................. (2)
Radical axis of (1) and (2) : (L) : 2x - 2y - 2 = 0
⇒ x - y - 1 = 0
Coaxial System of Circle : S + λL = 0
x2 + y2 + 4x + 2y + 5 + λ ( x - y - 1 ) = 0
⇒ x2 + y2 + ( 4 + λ ) x + ( 2 - λ ) y + ( 5 - λ ) = 0
⇒ 16 + λ2 + 8 λ + λ2 + 4 - 4 λ - 20 + 4 λ = 0
⇒ 2λ2 + 8λ = 0
⇒ λ2 + 4λ = 0
⇒ λ [λ + 4] = 0
⇒ λ = 0; λ = - 4
Limiting points : λ = 0 ⇒ (- 2, - 1)
λ = - 4 ⇒ (0, - 3)
Now any circle through the limiting points be x2 + y2 + 2gx + 2fy + c = 0
Where 4 +1 - 4g - 2f + c = 0 and 0 + 9 + 0 - 6f + c = 0
⇒ - 4g - 2f + c + 5 = 0 and - 6f + c + 9 = 0
Solving - 4g - 2f + c + 5 = 0
- 6f + c + 9 = 0
