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?

Solutions Collecting From Web of "Identify the plane defined by $|z-2i| = 2|z+3|$"

I assume this means locus and not plane..

$$ \begin{align*}
|z-2i|&= 2|z+3|\\
|x+iy-2i|&= 2|x+iy+3| \\
|x+i(y-2)|&= 2|(x+3)+i(y)| \\
x^2+y^2-4y+4&=4x^2+24x+36+4y^2 \\
x^2+8x+\frac{32}{3}+y^2+\frac{4}{3}y&=0 \\
x^2+8x+y^2+\frac{4}{3}y &= -\frac{32}{3}\\
x^2+8x+16 +y^2+\frac{4}{3}y +\frac{4}{9}&= -\frac{32}{3}+16+\frac{4}{9} \\

Which means our locus is a circle with centre $\left(-4,\frac{-2}{3}\right)$ and radius $\frac{2\sqrt{13}}{3}$.

You can square without adding spurious solutions, because $|w|\ge0$ by definition. So $|z-2i|^2=4|z+3|^2$ and, using $|w|^2=w\bar{w}$,
This simplifies to
Now recall that, for $z=x+yi$, $z\bar{z}=x^2+y^2$, $z+\bar{z}=2x$ and $z-\bar{z}=2yi$, to get
that can also be written as
or, completing the squares,
which represents a circle.

Your computation is also good up to
x^2+y^2-4y+4 = 4x^2+24x+36+4y^2
Then you do wrong simplifications: taking everything to the right hand side you get
exactly the same I got.

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 $\frac{2\sqrt{13}}{3}$