Let $\mathrm{z}$ 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

Given equation is

$\left|\frac{z-2i}{z+i}\right|=2, z\ne -i$

On simplifying the equation, we get

$(z-2i)(\overline{z}+2i)=4(z+i)(\overline{z}-i)$

$\Rightarrow \mathrm{z}\overline{\mathrm{z}}+4+2\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}})=4(\mathrm{z}\overline{\mathrm{z}}+1+\mathrm{i}(\overline{\mathrm{z}}-\mathrm{z}\left)\right)$

$\Rightarrow 3z\overline{z}-6i(z-\overline{z})=0$

Now putting the value of $z=x+iy \& \overline{z}=x-iy$ we get,

$\Rightarrow x^2+y^2-2i\left(2iy\right)=0$

$\Rightarrow x^2+y^2+4y=0$

On comparing the above equation with the general equation of the circle i.e

$(x-h{)}^2+(y-k{)}^2=r^2$

We get, centre $(0, -2)$