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
 A: Here's another approach altogether:
$$\begin{align}
|z-4i|=2|z+4|
&\iff\left|z-4i\over z+4 \right|=2\\
&\iff{z-4i\over z+4 }=2e^{i\theta}\quad\text{for some }\theta\in\mathbb{R}\\
&\iff z={8e^{i\theta}+4i\over1-2e^{i\theta}}
\end{align}$$
A: 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 $\frac{8\sqrt2}{3}$.
A: So far so good.
You may continue to see the geometry of the solution as well. 
$$3x^2+3y^2+32x+8y+48=0$$
$$x^2+y^2+(32/3)x+(8/3)y+16=0$$
Which is the equation of a circle. 
A: Your solution looks fine, but you should realize that it is an equation for the  circle with center $\left(-\dfrac{16}{3},-\dfrac{4}{3}\right)$ and with radius $\dfrac{8\sqrt{2}}{3}$.  If I were your grader, I would not give you a full credit for simply finding the final equation, yet not realizing it gives a circle.  Here is an alternative solution, using Euclidean geometry of the plane.
Let $A$ denote the point $4\text{i}$ of the complex plane $\mathbb{C}\cong\mathbb{R}^2$, whilst $B$ is the point $-4$.  Thus, if the point $C$ with complex coordinate $z$ satisfies
$$|z-4\text{i}|=2\,|z+4|\,,$$
then
$$CA=2\,CB\,.$$
On the line $AB$, there are two solutions $D$ and $E$, with complex coordinates
$$\frac{1}{3}\,(4\text{i})+\frac{2}{3}\,(-4)=\frac{-8+4\text{i}}{3}\text{ and }(-1)\,(4\text{i})+2\,(-4)=-8-4\text{i}\,,$$
respectively.  Thus, the point $C$ is a point such that $CD$ is the internal angular bisector of $\angle ACB$ and $CE$ is the external angular bisector of $\angle ACB$.  We can easily show that the locus of $C$ is a circle $\Gamma$ with diameter $DE$.
Thus, the center $P$ of $\Gamma$ has the complex coordinate
$$\frac{1}{2}\,\left(\frac{-8+4\text{i}}{3}\right)+\frac{1}{2}\,(-8-4\text{i})=\frac{-16-4\text{i}}{3}\,.$$
The radius of $\Gamma$ is $$\frac{1}{2}\,\Biggl|\left(\frac{-8+4\text{i}}{3}\right)-(-8-4\text{i})\Biggr|=\frac{8\sqrt{2}}{3}\,.$$
In other words, the complex coordinate $z$ of $C$ satisfies
$$\Biggl|z-\left(\frac{-16-4\text{i}}{3}\right)\Biggr|=\frac{8\sqrt{2}}{3}\,.$$
We can also write
$$z=\left(\frac{-16-4\text{i}}{3}\right)+\frac{8\sqrt{2}}{3}\,\exp(\text{i}\theta)\,,$$
where $\theta\in\mathbb{R}$.

In general, any solution $z\in\mathbb{C}$ to $|z-a|=r\,|z-b|$, where $r\in\mathbb{R}_{>0}\setminus\{1\}$ and $a,b\in\mathbb{C}$ is given by the circle
$$\left|z-c\right|=\rho\,.$$
Here, $c:=\dfrac{-a+r^2b}{r^2-1}$ and $\rho:=\dfrac{r}{|r^2-1|}\,|a-b|$.  In other words,
$$z=c+\rho\,\exp(\text{i}\theta)\,,$$
where $\theta\in\mathbb{R}$.  (Note that the solutions in the case where $r=1$ form a degenerate circle---a straight line.  This straight line is the perpendicular bisector of the segment joining $a$ and $b$.  It is given by the equation $(a-b)\,\bar{z}+(\bar{a}-\bar{b})z=|a|^2-|b|^2$, or equivalently, $\text{Re}\big((\bar{a}-\bar{b})\,z\big)=\dfrac{|a|^2-|b|^2}{2}$.  In other words, $z=\dfrac{a+b}{2}+\text{i}(a-b)\,t$, where $t\in\mathbb{R}$.)
A: Your computation is fine.
But I’ve been meditating on the form of your final result, since it says that the locus is a circle: coefficients of $x^2$ and $y^2$ are the same, and no $xy$ term. So there must be a theorem from plane geometry that says: If $P$ and $Q$ are points in the plane, the locus of points $R$ that are $k$ times as far from $Q$ as from $P$ is a circle, if $k\ne1$.
It’s easy enough to prove this analytically, by a method the same as yours, but surely there has to be a synthetic proof, perhaps even straight out of Euclid.
Does anyone know such a proof? Or can you cook one up? I looked at it, and have been stumped so far.
Anyway, thanks for a provocative question (and calculation).
