Question

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

Answer 1

Answer: 7

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