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

Let $(t, m)$ be the other end of the chord drawn from the point $(p, q)$ on the circle

$x^2+y^2=px+qy$

Their mid-point is $\left(\frac{t+p}{2},\frac{m+q}{2}\right)$

since, mid-point lies on $X$-axis i.e. $y=0$

$m+q=0$   $...\left(\mathrm{i}\right)$

Also, $(t, m)$ lies on the circle.

$\therefore t^2+m^2-pt-qm=0$   $...\left(\mathrm{ii}\right)$

From Eqs. $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get $t^2-pt+2q^2=0$

Which is quadratic in $t$ such that, Discriminant $>0$

$\Rightarrow p^2-8q^2>0\Rightarrow p^2>8q^2$