Let a point P be such that its distance from the point (5,0) is thrice the distance of P f…

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: 56.25

Accepted Answer

Let $P=(h,k)$. $PA^2=9PB^2$ with $A=(5,0),B=(-5,0)$:
$$(h-5)^2+k^2=9[(h+5)^2+k^2]\ \Longrightarrow\ 8h^2+8k^2+100h+200=0\ \Longrightarrow\ x^2+y^2+\frac{25}2x+25=0.$$

Centre $\left(-\frac{25}4,0\right)$, $r^2=\left(\frac{25}4\right)^2-25=\frac{625}{16}-25=\frac{225}{16}$.

$$4r^2=\frac{225}4=56.25.$$

Solutions · 1