Question

Let a point $P$ be such that its distance from the point $(5,0)$ is thrice the distance of $P$ from the point $(-5,0).$ If the locus of the point $P$ is a circle of radius $r,$ then $4r^2$ is equal to

Answer 1

Let $P$ be $(h,k),A(5,0)$ and $B(-5,0)$
Given $PA=3PB$
$\Rightarrow P{A}^2=9P{B}^2$
$\Rightarrow (h-5{)}^2+k^2=9\left[\right(h+5{)}^2+k^2]$
$\Rightarrow 8h^2+8k^2+100h+200=0$
$\therefore$ Locus of P is $x^2+y^2+\left(\frac{25}{2}\right)x+25=0$
Centre $\equiv (-\frac{25}{4},0)$
$\therefore r^2={\left(\frac{-25}{4}\right)}^2-25$
$=\frac{625}{16}-25=\frac{225}{16}$
$\therefore 4r^2=4\times \frac{225}{16}=\frac{225}{4}=56.25$