If $x-4=0$ is the radical axis of two orthogonal circles out of which one is $x^2+y^2=36$, then the centre…
If $x-4=0$ is the radical axis of two orthogonal circles out of which one is $x^2+y^2=36$, then the centre of the other circles is
$(8,0)$
$(9,0)$
$(6,0)$
$(12,0)$
Solution
Let equation of other circle be
$\begin{aligned}
& S^{\prime} \equiv x^2+y^2+2 g x+2 f y+c=0 \\
& S \equiv x^2+y^2=36
\end{aligned}$
$x^{\prime}-4=0 \text { is radicalaxis } \Rightarrow S^{\prime}=x^2+y^2-36+k(x-4)$ As circles are orthogonal
$\begin{aligned}
& \Rightarrow 2 g(0)+2 f(0)=4 \mathrm{k}-36-36 \Rightarrow k=-18 \\
& \therefore S^{\prime}=x^2+y^2-18 x-36=0
\end{aligned}$ So centre is $(9,0)$.