If two distinct chords, drawn from the point (p, q) on the circle $x^2+y^2=px+qy$, where $pq\ne 0$ are bisected

by the x-axis, then

The equation of the chord bisected at M(h, 0) is 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x{x}_1+y{y}_1-p\left(\frac{x+x_1}{2}\right)-q\left(\frac{y+y_1}{2}\right)\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=x_1^2+y_1^2-p{x}_1-q{y}_1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}hx+0-p\left(\frac{x+h}{2}\right)-q\left(\frac{y+0}{2}\right)\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=h^2+0-ph-0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2hx-px-ph-qy=2h^2-2ph\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2hx-px-qy=2h2-ph\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2h^2-2hx-ph+(px+qy)=0\end{array}$

which is passing through (p, q)

So, $2h^2-2ph-ph+\left(p^2+q^2\right)=0$

$\Rightarrow 2h^2-3ph+\left(p^2+q^2\right)=0$

Clearly, $D>0$

$\begin{array}{l}\Rightarrow 9p^2-8\left(p^2+q^2\right)>0\\ \Rightarrow p^2-8q^2>0\\ \Rightarrow p^2>8q^2\end{array}$