# Let a point P be such

Question:

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

Solution:

Let $P(h, k)$

Given

$P A=3 P B$

$P A^{2}=9 P B^{2}$

$\Rightarrow(h-5)^{2}+k^{2}=9\left[(h+5)^{2}+k^{2}\right]$

$\Rightarrow 8 h^{2}+8 k^{2}+100 h+200=0$

$\therefore$ Locus

$x^{2}+y^{2}+\left(\frac{25}{2}\right) x+25=0$

$\therefore c \equiv\left(\frac{-25}{4}, 0\right)$

$r^{2}=\left(\frac{-25}{4}\right)^{2}-25$

$=\frac{625}{16}-25$

$=\frac{225}{16}$

$\therefore 4 r^{2}=4 \times \frac{225}{16}=\frac{225}{4}=56.25$