Question

$P(-2,-1)$ and $0(0,3)$ are the limiting points of a coaxial system of which $C=x^2+y^2+5x+y+4=0$ is a member. The circle $S=x^2+y^2-4x-15=0$ is orthogonal to the circle $C=0$. The point where the polar of $P$ with respect to $C=0$ cuts the circle $S=0$ is

• A. $(3,6)$
• B. $(-3,6)$
• C. $(-6,3)$
• D. $(6,3)$

Answer 1

The correct option is D $(6,3)$

Pole of $P(-2,1)$ w.r.t to $c=x^2+y^2+5x+y+4=0$ is $T=0$
$-2x-y+\frac{5}{2}(x-2)+\frac{1}{2}(y-2)+4=0$
$\frac{x}{2}-\frac{y}{2}-5+4-\frac{1}{2}=0$
$x-y=3$
so,$s=x^2+y^2-4x-15=0$
so, put $y=x-3$ in $s=0$
$x^2+(x-3{)}^2-4x-15=0$
$2x^2-10x-6=0$
$x^2-5x-3=0$
$x=\frac{5_{-}^{+}\sqrt{25+12}}{2}$
$x=\frac{5\pm \sqrt{37}}{2}$ and $y=\frac{-1\pm \sqrt{37}}{2}$