Maximum area of Triangle $\Delta PAB$ In Complex plane $A$ and $B$ are two points given by $z_1=5-2i$ and $z_2=1+i$ and if $P(z)$ is any Point such that $$|z-z_1|=2|z-z_2|$$ Find the Maximum area of Triangle $\Delta PAB$.
I have done this problem algebraically using Heron's Formula taking $PA=2x$ and $PB=x$ and $AB=5$ . Is there any geometrical way of doing this? 

First, the area of a triangle $\Delta PAB$ is unchanged by translations or rotations of the complex plane. Since $|AB|=5$, we may therefore choose coordinates so that the points $A$ and $B$ are instead at $0$ and $5$ respectively. Writing the third point as $z=x+I y$, the area of $\Delta PAB$ is then $\frac{5}{2}y$ since the base and height are $5$ and $y$ respectively, and our goal becomes to maximize $y$.
To do so, note that the condition for $P(z)$ may be expressed as
$$|z|=2|z-5|\implies x^2+y^2=4(x-5)^2+4y^2\implies \left(x-\frac{20}{3}\right)^2+y^2=\left(\frac{10}{3}\right)^2$$
where we have squared both sides and then completed the squares. Thus $y^2=\left(\frac{10}{3}\right)^2-\left(x-\frac{20}{3}\right)^2\leq \left(\frac{10}{3}\right)^2,$ so $|y|\leq \frac{10}{3}$ (with equality iff $x=\frac{20}{3}$) and we conclude that $\Delta PAB$ has an area of at most $\frac{5}{2}\cdot \frac{10}{3}=\frac{25}{3}$.
