Geometric interpretation of a complex set

These usually aren’t too bad but I had difficulties thinking of what the set
$$\{z\in\mathbb{C}:|z+i|=2|z|\}$$
looks like in the complex plane. I got as far as
$$|z+i|=2|z|\Rightarrow \sqrt{(z+i)(\overline{z+i})}=2z\overline{z}\Rightarrow
\sqrt{z\overline{z}+i\overline{z}-iz+1}=2z\overline{z}\\
\Rightarrow z\overline{z}-i(z+
z\overline{z})+1=4z\overline{z}\Rightarrow 3z\overline{z}=1-i2\Re(z)$$

And rearranging a bit I guess I can see that $z\overline{z}+\frac{2i\Re(z)}{3}=1/3$. If there weren’t that second term on the left, this would be clear, but I am not sure how to think about the dependence on the real part of $z$ here.

Having graphed in wolfram, I know this is just a circle of radius 2/3 centered at $i/3$, but I would appreciate help with the thought process.

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Using the formula
$$
\lvert z+w \rvert^2 = \lvert z \rvert^2 + 2 \Re(z \overline w)+ \lvert w \rvert^2
$$
you have
$$
\lvert z+i \rvert = 2 \lvert z \rvert \\
\Longleftrightarrow \lvert z \rvert^2 + 2 \Re(-iz) + 1 = 4 \lvert z \rvert^2 \\
\Longleftrightarrow \lvert z \rvert^2 – \frac 23 \Re(-iz) = \frac 13
$$
Now “complete the square”:
$$
\Longleftrightarrow \lvert z \rvert^2 – 2 \Re(- \frac i3 z ) + \lvert \frac i3 \rvert^2 = \frac 13 + \frac 19 \\
\Longleftrightarrow \lvert z – \frac i3 \rvert^2 = \frac 49
$$

If $z=x+iy$ where $x,y$ are real

$$(y+1)^2+x^2=4(x^2+y^2)\iff3x^2+3y^2-2y-1=0$$

$$\iff(3y-1)^2+(3x)^2=4$$ which is a circle centered at $(?,?)$ with the radius $=?$