Which $z\in\mathbb{C}$ satisfy the equation $|z-i|=\sqrt{2}|\bar{z}+1|$? Let $S$ be the set of all complex numbers $z$ satisfying the rule $$|z-i|=\sqrt{2}|\bar{z}+1|$$
Show that $S$ contains points on a circle.
My attempt,
By substituting $z = x + yi$, and squaring both sides. But I can't get the circle form. 

Let $z=x+iy$. Note that $\overline{z}=x-iy$.
$$\begin{align}
|z-i|&=\sqrt{2}|\overline{z}+1|\\
|z-i|^2&=2|\overline{z}+1|^2\\
|x+i(y-1)|^2&=2|(x+1)-iy|^2\\
x^2+(y-1)^2&=2(x+1)^2+y^2)\\
x^2+4x+2+y^2+2y-1&=0\\
(x+2)^2+(y+1)^2&=2^2
\end{align}$$
