Mathematics - Circle Question with Solution | TestHub

Let the roots of $2x^2-rx+p=0$ are $x_1, x_2$ and roots of $y^2-sy-q=0$ are $y_1, y_2$

So, $x_1+x_2=\frac{r}{2}, x_1{x}_2=\frac{p}{2}, y_1+y_2=s, y_1{y}_2=-q$

Equation of the circle with $PQ$ as diameter will be 

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$ 

i.e. $2\left(x^2+y^2\right)-rx-2sy+p-2q=0$ 

On comparing with the given equation $r=11,s=7$

$p-2q=-22$

$\therefore 2r+s-2q+p=22+7-22=7$