Let the equation of required circle be, x2+y2+2gx+2fy+c=0 ....(I) The above passes through point (3, 2) so, 9+4+6g+4f+c=0 6g+4f+c+13=0 The equation (I) bisects the circumference of circle x2+y2=15 So, the common chord passes through the center of the circle, x2+y2=15 So, c+15=0 c=−15 Since equation (I) cuts the circle, x2+y2+4x+6y+3=0 Orthogonally so, 4g+6f=c+3 From above equations, g=3,f=−4,c=−15 So, the equation of the required circle is, x2+y2+6x−8y−15=0