Let, $z=x+iy$
$\mathrm{I}\mathrm{m}\left(\frac{i\left(x+yi\right)+2}{x+iy+i}\right)=-1$
$\Rightarrow \mathrm{I}\mathrm{m}\left(\frac{-\left(y-2\right)+ix}{x+i\left(y+1\right)}\right)=-1$
$\Rightarrow \mathrm{I}\mathrm{m}\left(\frac{\left(-\left(y-2\right)+ix\right)\left(x-i\left(y+1\right)\right)}{x^2+{\left(y+1\right)}^2}\right)=-1$
$\Rightarrow x^2+y^2-y-2=-x^2-y^2-2y-1$
$\Rightarrow 2x^2+2y^2+y-1=0$
$\Rightarrow x^2+y^2+\frac{y}{2}-\frac{1}{2}=0$
$\Rightarrow$ the radius $\left(r\right)$ of the circle is $\sqrt{{\left(\frac{1}{4}\right)}^2+\frac{1}{2}}=\frac{3}{4}$

Hence, $4r^2=4\times \frac{9}{16}=\frac{9}{4}$$=2.25$