The polar of $P\left(3,5\right)$ with respect to circles $x^2+y^2-16x+36=0$ and $x^2+y^2+16x+36=0$ intersect at $Q$. Then the circle with $PQ$ as diameter passes through

The polar of the point $P\left(3,5\right)$ with respect to the circle $x^2+y^2-16x+36=0$ is  is $\begin{array}{rcl}3x+5y-8\left(x+3\right)+36& =& 0\\ -5x+5y+12& =& 0\\ 5x-5y-12& =& 0\end{array}$

The polar of the point $P\left(3,5\right)$ with respect to the circle $x^2+y+16x+36=0$

$\begin{array}{rcl}3x+5y+8\left(x+3\right)+36& =& 0\\ 11x+5y+60& =& 0\end{array}$

The point of intersection of the above two lines is as below

$\begin{array}{rcl}5x-5y+11x+5y-12+60& =& 0\\ 16x& =& -48\\ x& =& -3\end{array}$

when $x=-3,$the value of $y$ is $y=-\frac{27}{5}$

Equation of the circle having $P\left(3,5\right)$ and $\left(-3,-\frac{27}{5}\right)$ as ends of diameter

$\left(x-3\right)\left(x+3\right)+\left(y-5\right)\left(y+\frac{27}{5}\right)=0$

when $y=0$

$\begin{array}{rcl}\left(x-3\right)\left(x+3\right)& =& 27\\ x^2& =& 36\\ x& =& \pm 6\end{array}$

Hence, the points are $\left(\pm 6,0\right)$