Question

Two distinct chords drawn from the point $(p,q)onthecircle$x^{2} + y^{2} = px + qy$,where$pq \neq 0$ are bisected by the x-axis. Then

• A. $|p|=|q|$
• B. $p^2=8q^2$
• C. $p^2<8q^2$
• D. $p^2>8q^2$

Answer 1

The correct option is D $p^2>8q^2$

$AB$ is chord, $M(h,0)$ is mid point

$B=(-p+2h,-q)$ put in $x^2+y^2-px-qy=0$, we get

$2h^2-3ph+(p^2+q^2)=0$

Since, there are two chords, $D>0$

$\Rightarrow {(3-p)}^2-4\left(2\right)(p^2+q^2)>0$ $\Rightarrow p^2>8q^2$