Question

Let the abscissae of the two points $P$ and $Q$ be the roots of $2 x ^{2}- rx + p =0$ and the ordinates of $P$ and $Q$ be the roots of $x^{2}-s x-q=0$. If the equation of the circle described on $PQ$ as diameter is $2\left( x ^{2}+ y ^{2}\right)-11 x -14 y -22=0$, then $2 r+s-2 q+p$ is equal to

Answer 1

Equation of the circle with $PQ$ as diameter is
$2\left(x^{2}+y^{2}\right)-r x-2 s y+p-2 q=0$
on comparing with the given equation
$r=11, s=7 $
$p-2 q=-22$
$\therefore 2 r+s-2 q+p=22+7-22=7$