If a circle $S$ passing through the point $(3,4)$ cuts the circle $x^2+y^2=36$ orthogonally, then the locus…

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

Solution

Let the circle is $x^2+y^2+2 g x+2 f y+c=0$, having centre $(-g,-f)$, since it passes through the point $(3,4)$
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)

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