AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Let the equation of the circle is,
x2+y2+2gx+2gy+c=0
have center (−g,−f) . As it passes through the point (3, 4)
So,
9+16+6g+8f=0
The circle is intersecting another circle (x2+y2=36) orthogonally,
2g(0)+f(0)−c−26
c=36
From above calculations,
−6g−8f=61
By taking locus of the point (−g,−f) , we get 6x+8y−61=0