AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

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