Given equation is
∣z+iz−2i∣=2,z=−i
On simplifying the equation, we get
(z−2i)(zˉ+2i)=4(z+i)(zˉ−i)
⇒zzˉ+4+2i(z−zˉ)=4(zzˉ+1+i(zˉ−z))
⇒3zzˉ−6i(z−zˉ)=0
Now putting the value of z=x+iy&\bar{z}=x-iy we get,
⇒x2+y2−2i(2iy)=0
⇒x2+y2+4y=0
On comparing the above equation with the general equation of the circle i.e
(x−h)2+(y−k)2=r2
We get, centre (0,−2)