Question

Let $x-4=0$ be the radical axis of two circles which are intersecting orthogonally. If $x^{2}+y^{2}=36$ is one of those circles, then the other circle is

• A. $x^{2}+y^{2}-16 x+36=0$
• B. $x^{2}+y^{2}-18 x+36=0$
• C. $x^{2}+y^{2}-18 x+24=0$
• D. $x^{2}+y^{2}-6 x+8 y+36=0$

Answer 1

Answer: (B)

We have, equation of one circle
$x^{2}+y^{2}=36 $
$\Rightarrow \, x^{2}+y^{2}-36 =0\,\,\,\,\,\,\,\dots(i)$
and radical axis of two circle is $x-4=0$
So, equation of other circle is
$x^{2}+y^{2}-36+k(x-4)=0$
$x^{2}+y^{2}+k x-4 k-36=0\,\,\,\,\,\,\dots(ii)$
Both circles are intersecting orthogonally, then
$-4 k -36-36=0 $
$-4 k =72 $
$k =-18$
So, equation of required circle
$x^{2}+y^{2}-18 x-4(-18)-36=0$
$\Rightarrow \,x^{2}+y^{2}-18 x+72-36=0$
$\Rightarrow \, x^{2}+y^{2}-18 x+36=0$