Question

If the circle passes through the point $(a, b)$ and cuts the circle $x^{2}+y^{2}=k^{2}$ orthogonally, then the locus of its centre is given by:

• A. $2 a x+2 b y-\left(a^{2}+b^{2}+k^{2}\right)=0$
• B. $2 a x+2 b y+\left(a^{2}+b^{2}-k^{2}\right)=0$
• C. $2 a x+2 b y+\left(a^{2}+b^{2}+k^{2}\right)=0$
• D. None of the above

Answer 1

Answer: (A)

If $S_{1}=x^{2}+y^{2}+2 g_{1} x+2 f_{1} y+c_{1}=0$
and $S_{2}=x^{2}+y^{2}+2 g_{2} x+2 f_{2} y+c_{2}=0$
are two circles orthogonal to each other, if
$2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2}$
Let the equation of circle be
$x^{2}+y^{2}+2 g x+2 f y+c=0\,\,\,\,\,\dots(i)$
Its centre is $(-g,-f)$.
This is orthogonal to given circle
$ x^{2}+y^{2} =k^{2} $
$\Rightarrow \, g(0)+f(0) =c-k^{2} $
$\Rightarrow \, c=k^{2}$
$\therefore $ From Eq. (i)
$x^{2}+y^{2}+2 g x+2 f y+k^{2}=0$
Also, this circle passes through $(a, b)$
$\therefore \, a^{2}+b^{2}+2 g a+2 f b+k^{2}=0$
Locus of centre $(-g,-f)$ is
$a^{2}+b^{2}-2 x a-2 y b+k^{2}=0$
$\Rightarrow \, 2 a x+2 b y-a^{2}-b^{2}-k^{2}=0$