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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
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Verified Answer
The correct answer is:
$6 x+8 y-61=0$
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 $6 x+8 y-61=0$.
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 $6 x+8 y-61=0$.
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