Let the lines 2 - i z = 2 + i z ¯ and 2 + i z + i - 2 z ¯ - 4 i = 0 , (here i 2 = - 1 ) be normal…

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

$\Rightarrow (2-i)(x+iy)=(2+i)(x-iy)$

$\Rightarrow 2x-ix+2iy+y=2x+ix-2iy+y$

$\Rightarrow 2ix-4iy=0$

$L_1:x-2y=0$

$\Rightarrow (2+i)z+(i-2)\overline{z}-4i=0$

$\Rightarrow (2+i)(x+iy)+(i-2)(x-iy)-4i=0$

.$\Rightarrow 2x+ix+2iy-y+ix-2x+y+2iy-4i=0$

$\Rightarrow 2ix+4iy-4i=0$

Solve $L_1$ and $L_2, 4y=2, y=\frac{1}{2}$

$\therefore x=1$

Centre $\left(1,\frac{1}{2}\right)$

$L_3:iz+\overline{z}+1+i=0$

$\Rightarrow i(x+iy)+x-iy+1+i=0$

$\Rightarrow ix-y+x-iy+1+i=0$

$\Rightarrow (x-y+1)+i(x-y+1)=0$

Radius $=$ distance from $\left(1, \frac{1}{2}\right)$ to $x-y+1=0$

$r=\left|\frac{1-\frac{1}{2}+1}{\sqrt{2}}\right|$

$r=\frac{3}{2\sqrt{2}}$