Question

Find the locus of $P$ for which the distance from $P$ to origin is double the distance from $P$ to the point $(1,2)$.

Answer 1

Let the point is $P(h,k)$

Distance of $P$ from origin $=\sqrt{(h-0{)}^2+(k-0{)}^2}$

$=\sqrt{h^2+k^2}$

Distance of $P$ from the point $(1,2)$ $=\sqrt{(h-1{)}^2+(k-2{)}^2}$

$=\sqrt{h^2+1-2h+k^2+4-4k}$

$=\sqrt{h^2-2h+k^2-4k+5}$

Now,

$\sqrt{h^2+k^2}$$=2\sqrt{h^2-2h+k^2-4k+5}$

Squaring both sides,

$h^2+k^2=4(h^2-2h+k^2-4k+5)$

$h^2+k^2=4h^2+4k^2-8h-16k+20$

$3h^2+3k^2-8h-16k+20=0$

So, the locus of $P$ is

$3x^2+3y^2-8x-16y+20=0$