If a circle $S$ passing through the point $(3,4)$ cuts the circle $x^2+y^2=36$ orthogonally, then the locus…
- $x^2+y^2-6 x-8 y+11=0$
- $6 x+8 y-61=0$
- $x^2+y^2-8 x-6 y+11=0$
- $6 x+8 y+11=0$
Solution

And circle is intersecting the other circle $ \begin{aligned} & x^2+y^2=36 \text { orthogonally, so } \\ & \qquad 2 g(0)+2 f(0)=c-36 \end{aligned} $

From Eqs. (i) and (ii) $ -6 g-8 f=61, $ Now, on taking locus of point $(-g,-f)$, we are getting $6 x+8 y-61=0$
Asked in: AP EAMCET 2018 (22 Apr Shift 2)