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 $2(x^2+y^2)–11x–14y–22=0$, then $2r+s–2q+p$ is equal to .

Answer 1

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$
$\therefore$ Equation of circle $\equiv (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$
$\Rightarrow x^2-(x_1+x_2)x+x_1{x}_2+y^2-(y_1+y_2)y+y_1{y}_2=0$
$\Rightarrow x^2-\frac{r}{2}x+\frac{p}{2}+y^2-sy-q=0$
$\Rightarrow 2x^2+2y^2-rx-2sy+p-2q=0$
Compare with $2x^2+2y^2-11x-14y-22=0$
We get $r=11,s=7,p-2q=-22$
$\Rightarrow 2r+s+p-2q=22+7-22=7$