The equation of the circle which passes through the point (3,2) bisects the circumference of the circle x^2+y^2=15 and cuts the circle x^2+y^2+4 x+6 y+3=0 orthogonally is

Step-by-step solution

Let the equation of required circle is x^2+y^2+2 g x+2 f y+c=0 Since, circle (i) passes through point (3,2) , so aligned & & 9+4+6 g+4 f+c & =0 & & 6 g+4 f+c+13 & =0 aligned Since, circle (i) bisects the circumference of the circle x^2+y^2=15 , so the common chord passes through the centre of the circle x^2+y^2=15 . So, aligned c+15 & =0 c & =-15 aligned Since, circle Eq. (i) cuts the circle x^2+y^2+4 x+6 y+3=0 Orthogonally, so 4 g+6 f=c+3 From Eqs. (ii), (iii) and (iv) g=3, f=-4, c=-15 So, required equation of cir