The equation of a circle passing through the points of intersection of the circles x2+y2 - 4x - 6y - 12 = 0 , x2+y2 + 6x + 4y - 12 = 0 and having radius √13 is
The equation of a circle passing through the points of intersection of the circles, (x2+y2−4x−6x−12)+λ(x2+y2+6x+4y−12)=0 x2(1+λ)+y2(1+λ)+x(6λ−4)+y(4λ−4)−12λ−12=0 Rewrite the above equation. x2+y2+‌
x(6λ−4)
(1+λ)
+‌
y(4λ−6)
(1+λ)
−12=0 On comparing in equations with x2+y2+2gx+2fy+c=0 g=‌
(3λ−2)
(1+λ)
,f=‌
(2λ−3)
(1+λ)
,c=−12 It is given that the radius is √13 13=‌
(3λ−2)2
(1+λ)2
+‌
(2λ−3)2
(1+λ)2
+12 13(1+λ)2=(3λ−2)2+(2λ−3)2+12(1+λ)2 After simplification, 13λ2+13+26λ=25λ2+25 12λ2−26λ+12=0 6λ2−13λ+6=0 (2λ−3)(3λ−2)=0 λ=‌
3
2
,‌
2
3
The required equation when λ=‌
3
2
x2(1+‌
3
2
)+y2(1+‌
3
2
)+x(6×‌
3
2
−4)+y(4×‌
3
2
−6)−12×‌
3
2
−12=0 Further simplification, x2+y2+2x−12=0 The required equation when λ=‌