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

$$\begin{gathered} x^2+y^2-36+k(x-4)=0 \\ {x}^2+y^2+kx-4k-36=0 \end{gathered}$$
Both circles are intersecting orthogonally then
$\begin{array}{l}-4k-36-36=0\\ -4k=72,\Rightarrow k=-18\end{array}$
So, equation of required circle
$\begin{array}{l}{x}^2+y^2-18x-4(-18)-36=0\\ x^2+y^2-18x+72-36=0\\ x^2+y^2-18x+36=0\end{array}$