$$\begin{gathered} If two distinct chords, drawn from the point (p,q) on the \\ circle x^2+y^2-px-qy=0 (where p q\ne 0) are bisected by the x-axis then \end{gathered}$$

$$\begin{gathered} Let A=(p,q) and AB is chord bisected by x-axis at M(h,0) \\ \therefore B=(2h-p,-q) lies on x^2+y^2-px-qy=0 \\ \Rightarrow (2h-p{)}^2+(-q{)}^2-p(2h-p)-q(-q)=0 \\ \Rightarrow 4h^2-4hp+p^2+q^2-2ph+p^2+q^2=0 \\ \Rightarrow 4h^2-6ph+2p^2+2q^2=0 \\ \Rightarrow Here \text{'} 2 \text{'} chords meet x-axis at two points \\ \Rightarrow \therefore \Delta >0 \\ \Rightarrow 36p^2-4\times 4\times 2p^2+q^2>0 \\ \Rightarrow 4p^2-32q^2>0 \\ \therefore p^2>8q^2 \end{gathered}$$