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

$\left|\frac{z-2 i}{z+i}\right|=2$ <br/><br/> $\Rightarrow (z-2 i)(\bar{z}+2 i)=4(z+i)(\bar{z}-i)$ <br/><br/> $\Rightarrow z \bar{z}+2 i z-2 i \bar{z}+4=4(z \bar{z}-z i+\overline{z i}+1)$ <br/><br/> $\Rightarrow 3 z \bar{z}-6 i z+6 i \bar{z}=0$ <br/><br/> $\Rightarrow z \bar{z}-2 i z+2 i \bar{z}=0$ <br/><br/> $\therefore$ Centre $(-2 i)$ or $(0,-2)$ <br/><br/><b>Other Method :</b> <br/><br/>$\left|\frac{z-2 i}{z+i}\right|=2, z \neq-i$ <br/><br/>Put $z=x+i y$ <br/><br/>$$ \begin{aligned} & \left|\frac{x+i y-2 i}{x+i y+i}\right|=2 \\\\ & \Rightarrow \left|\frac{x+i(y-2)}{x+i(y+1)}\right|^2=4 \\\\ & \Rightarrow x^2+(y-2)^2=4\left[x^2+(y+1)^2\right] \\\\ & \Rightarrow x^2+y^2+4-4 y=4\left[x^2+y^2+1+2 y\right] \end{aligned} $$ <br/><br/>$\Rightarrow$ $x^2+y^2+4 y=0$ <br/><br/>or $x^2+(y+2)^2=2^2$ <br/><br/>$\therefore$ Centre is $(0,-2)$.