Let z be a complex number such that  $\left|\frac{z-2i}{z+i}\right|=2,z\ne -i$. Then z lies on the circle of radius 2 and centre

$$\begin{gathered} (z-2i)(\overline{z}+2i)=4(z+i)(\overline{z}-i) \\ z\overline{z}+4+2i(z-\overline{z})=4(z\overline{z}+1+i(\overline{z}-z)) \\ 3z\overline{z}-6i(z-\overline{z})=0 \\ {x}^2+y^2-2i\left(2iy\right)=0 \\ {x}^2+y^2+4y=0 \end{gathered}$$