$A(-3,1), B(0,-5), P(X,Y)$
If $|AP| = 2|BP|$ prove that $x$ and $y$ satisfy the equation:
\begin{aligned} \ x^2+y^2-2x+14y+30 =0 \end{aligned}
I get as far as determining the co-ordinates like so
\begin{aligned} \ \sqrt{(x+3)^2+(y-1)^2}= 2\sqrt{(x)^2+(y+5)^2} \end{aligned}
To
\begin{aligned} \ x^2+y^2+2y+40-6x =0 \end{aligned}
Which gives me $(3, -1)$, this won't satisfy, is my method correct?
If we start from the equation $$ \sqrt{(x+3)^2+(y-1)^2}= 2\sqrt{(x)^2+(y+5)^2}, $$ and square both sides, we obtain the equivalent equation $$x^2+6x+9+y^2-2y+1=4x^2+4y^2+40y+100,$$ which simplifies to $$3x^2+3y^2-6x+42y+90=0,$$ which is equivalent to $$x^2+y^2-2x+14y+30=0.$$
Comment: Draw the line segment that joins $A$ to $B$. It is fairly easy to see that $(-1,-3)$ is the point on this line segment that is twice as far from $A$ as it is from $B$. So $(-1,-3)$ ought to be on our circle, and indeed it is. That provides a partial check on the correctness of our computations.