The locus of the point from which mutually perpendicular tangents can be drawn to the circle $x^2+y^2=36$  is 

$y=mx\pm 6\sqrt{1+m^2}$ is the equation of  

any tangent to the circle. If it passes through$(h, k)$ , then

$\begin{array}{l}k=mh\pm 6\sqrt{1+m^2}\\ \Rightarrow (k-mh{)}^2=36\left(1+m^2\right)\\ \Rightarrow \left(36-h^2\right)m^2+2mhk+\left(36-k^2\right)=0\end{array}$

  which gives two values of m say m1 and m2, slopes of

 two tangents passing thought $(h, k).$These tangents are perpendicular 

If $m_1{m}_2=-1\Rightarrow \frac{36-k^2}{36-h^2}=-1$

$\Rightarrow h^2+k^2=72\Rightarrow$ then of $(h,k)$ is $x^2+y^2=72$