<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] I assume this means locus and not plane..
$$ \begin{align*}
|z-2i|&= 2|z+3|\\
|x+iy-2i|&= 2|x+iy+3| \\
|x+i(y-2)|&= 2|(x+3)+i(y)| \\
\sqrt{x^2+(y-2)^2}&=2\sqrt{(x+3)^2+y^2}\\
x^2+(y-2)^2&=4\left(x^2+6x+9+y^2\right)\\
x^2+y^2-4y+4&=4x^2+24x+36+4y^2 \\
3x^2+24x+32+3y^2+4y&=0\\
x^2+8x+\frac{32}{3}+y^2+\frac{4}{3}y&=0 \\
x^2+8x+y^2+\frac{4}{3}y &= -\frac{32}{3}\\ 
x^2+8x+16 +y^2+\frac{4}{3}y +\frac{4}{9}&= -\frac{32}{3}+16+\frac{4}{9} \\
(x+4)^2+\left(y+\frac{2}{3}\right)^2&=\frac{52}{9}
\end{align*}$$
Which means our locus is a circle with centre $\left(-4,\frac{-2}{3}\right)$ and radius $\frac{2\sqrt{13}}{3}$.
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