How can we prove the locus is a circle? Given two fixed points A and B, find the locus of the point P, satisfying PA=2PB. Of course we can use  Cartesian geometry to find the equation of the curve. 
Let the midpoint of A and B be the origin, and the line AB be the x-axis, then the coordinates of A and B is (a,0), (-a,0). Let the coordinates of P be (x,y). Then 
$\sqrt{(x-a)^2+y^2}=2\sqrt{(x+a)^2+y^2}$,
we get $(x+\frac{5}{3})^2+y^2=(\frac{4}{3}a)^2$. So the locus is a circle. But how can we solve this problem by pure geometry? 
 A: Without loss of generality, let the distance between $A$ and $B$ equal 3 (this is for convenience in calculations) and suppose that $\overline{AB}$ is horizontal (this just makes it easier to talk about where things are).

Now, the point $P_1$ that is on $\overline{AB}$ a distance of 2 from $A$ and 1 from $B$ is in the locus, as are points $P_2$ and $P_3$ that are directly above and below $B$, a distance of $\sqrt{3}$ from $B$ and $2\sqrt{3}$ from $A$.  Using these three points, we can determine that if the locus were a circle, its center would have to be at a point we'll call $C$, on $\overrightarrow{AB}$ a distance of 4 from $A$ and 1 from $B$ (1 unit past $B$ from $A$).

Next, consider a point $P$ in the locus with $PA=2x$, $PB=x$, and $PC=y$.  Apply Stewart's Theorem to get $$\begin{align}
3y^2+(2x)^2&=4(x^2+3) \\
3y^2+4x^2&=4x^2+12 \\
y^2=4 \\
y=2.
\end{align}$$  So, all possible points $P$ in the locus are a distance of 2 from $C$, which means that the locus is a circle of radius 2, centered at $C$.
n.b. This suggests that if $A=(a,0)$ and $B=(-a,0)$, $C=(-2a,0)$ and the radius is $2a$, which would give an equation of $(x+2a)^2+y^2=4a^2$, which is different from what you said you got.
