Question

If the locus of a points on which the line joining $(-5,1)$ and $(3,2)$ subtends a right angle, is given by $x^2+y^2+2x-3y-k=0$, where $k$ is a positive integer, then find the value of $k-6$.

Answer 1

Let $P(h,k)$ be a moving point and let $A(-5,1)$ and $B(3,2)$ be given points. By the given condition $\angle APB={90}^{\circ }$
$\therefore \Delta PAB$ is a right angled triangle
$\Rightarrow A{B}^2=A{P}^2+P{B}^2$
$\Rightarrow {(3+5)}^2+{(2-1)}^2={(h+5)}^2+{(k-1)}^2+{(h-3)}^2+{(k-2)}^2$
$\Rightarrow 65=2(h^2+k^2+2h-3k)+39$
$\Rightarrow h^2+k^2+2h-3k-13=0$
Hence locus of $(h,k)$ is $\Rightarrow h^2+k^2+2h-3k-13=0$

$\therefore k=13\Rightarrow k-6=7$