Question

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

• A. $x^{2}+y^{2}-6 x-8 y+11=0$
• B. $6 x+8 y-61=0$
• C. $x^{2}+y^{2}-8 x-6 y+11=0$
• D. $6 x+8 y+11=0$

Answer 1

Answer: (B)

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 $(3,4)$
So, $9+16+6 g+8 f + c=0$ ...(i)
And circle is intersecting the other circle
So $x^{2}+y^{2}=36$ orthogonally, so
$2 g(0)+2 f(0) =c-36$
$\Rightarrow C =36$ ...(ii)
From Eqs. (i) and (ii)
$-6 g-8 f=61$
Now, on taking locus of point $(-g,-f)$,
we are getting $6 x+8 y-61=0$