Question

If the abscissaes and ordinates of two points $P$ and $Q$ are the roots of the equations $x^2+2ax-b^2=0$ and $x^2+2px-q^2=0$ respectively, then equation of the circle with $PQ$ as diameter is

• A. $x^2+y^2+2ax+2py-b^2-q^2=0$
• B. $x^2+y^2-2ax-2py+b^2+q^2=0$
• C. $x^2+y^2-2ax-2py-b^2-q^2=0$
• D. $x^2+y^2+2ax+2py+b^2+q^2=0$

Answer 1

The correct option is A $x^2+y^2+2ax+2py-b^2-q^2=0$

Let $x_1,x_2$ and $y_1,y_2$ be the roots of $x^2+2ax-b^2=0$ and $x^2+2px-q^2=0$ respectively. Then
$x_1+x_2=-2a,x_1{x}_2=-b^2$ and $y_1+y_2=-2p,y_1{y}_2=-q^2$
$\therefore$ Equation of the circle with $P(x_1,y_1)$ and $Q(x_2,y_2)$ as the extremities of the diameter then
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\Rightarrow x^2+y^2-x(x_1+x_2)-y(y_1+y_2)+x_1{x}_2+y_1{y}_2=0\Rightarrow x^2+y^2+2ax+2py-b^2-q^2=0$