Find the equation of the locus

Question

There's a diagram that shows a straight line $AB$ with $A=(-3,-3)$ and $B=(3,6)$.

Question: Point $P$ moves such that its distance from $A$ is always twice its distance from $B$.

Find the equation of the locus of $P$.

How to solve this?

Answer 1

Let $P=(x,y)$, then

\begin{align*} \sqrt{(x+3)^{2}+(y+3)^{2}} &= 2\sqrt{(x-3)^{2}+(y-6)^{2}} \\ (x+3)^{2}+(y+3)^{2} &= 4[(x-3)^{2}+(y-6)^{2}] \\ x^2+y^2+6x+6y+18 &= 4(x^2+y^2-6x-12y+45) \\ 0 &= 3x^2+3y^2-30x-54y+162 \\ x^2+y^2-10x-18y+54 &=0\\ (x-5)^2+(y-9)^2&=52 \end{align*}

which is a circle with centre $(5,9)$ and radius $2\sqrt{13}$.