<s>[INST] Identify the plane defined by $|z-2i| = 2|z+3|$ I tried:
$$|z-2i| = 2|z+3| \Leftrightarrow \\
|x+yi-2i|=2|x+yi+2|\Leftrightarrow \\
\sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\
\sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\
y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\
-3y^2-4y=5x^2+24x+26 \Leftrightarrow \\
???$$
What do I do next?
 [/INST] Naive geometric solution:
The locus of points whose distances are a fixed ratio (not equality) from two given points is a circle. Call the points $A$ and $B$. The circle's center lies on the (extended) line $AB$.
If you want all the points twice as far from $A$ as from $B$, then the two points where its circumference intersects line $AB$ are $\frac13$ of the way from $B$ to $A$, and at distance $d(AB)$ on the other side of $B$ (opposite from $A$). The center is the midpoint between these points, at distance $\frac13 d(AB)$ on the far side of $B$ from $A$. That makes the radius of the circle $\frac{2}{3}d(AB)$.
In this case, $A$ is at $2i$ and $B$ is at $-3$. That puts the center at the point $2i(1-t)-3t$ for $t=\frac{4}{3}$, and it makes the radius $\frac{2}{3}|2i+3|$. I.e., the center is at $-4-\frac23i$, and the radius is $\frac{2\sqrt{13}}{3}$
</s>