<s>[INST] Solving $|z-4i|=2|z+4|$ $$\begin{align}
|z-4i|&=2|z+4|\\[4pt]
|x+yi-4i|&=2|x+yi+4|\\[4pt]
|x+i(y-4)|&=2|(x+4)+iy|\\[4pt]
\sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt]
(\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt]
x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt]
x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt]
0&= 3x^2+3y^2+32x+8y+48
\end{align}$$
Is it okay? Thank you
 [/INST] Your argument is completely fine. Continue it with:
$$x^2+\frac{32}{3}x+y^2+\frac 83 y+16=0$$
Now complete the square on the $x$'s and $y$'s
$$(x+\frac{16}{3})^2-\frac{256}{9}+(y+\frac43)^2-\frac{16}{9}+16=0$$
$$(x+\frac{16}{3})^2+(y+\frac{4}{3})^2=\frac{128}{9}$$
Thus we have a circle, centre $(-\frac{16}{3}, -\frac 43)$, radius $\frac{8\sqrt2}{3}$.
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