<s>[INST] Circle locus, how to satisfy the equation. 
$A(-3,1), B(0,-5), P(X,Y)$
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?
 [/INST] 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. 
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