Let the lines $(2-\mathrm{i}) \mathrm{z}=(2+\mathrm{i}) \overline{\mathrm{z}}$ and $(2+\mathrm{i}) \mathrm{z}+(\mathrm{i}-2) \overline{\mathrm{z}}-4 \mathrm{i}=0$, (here $\mathrm{i}^{2}=-1$ ) be normal to a circle $\mathrm{C}$. If the line $\mathrm{iz}+\overline{\mathrm{z}}+1+\mathrm{i}=0$ is tangent to this circle $\mathrm{C}$, then its radius is :
Correct Option: , 3
$(2-i) z=(2+i) \bar{z}$
$\Rightarrow(2-i)(x+i y)=(2+i)(x-i y)$
$\Rightarrow 2 x-i x+2 i y+y=2 x+i x-2-i y+y$
$\Rightarrow 2 i x-4 i y=0$
$L_{1}: x-2 y=0$
$\Rightarrow(2+i) z+(i-2) \bar{z}-4 i=0$
$\Rightarrow(2+i)(x+i y)+(i-2)(x-i y)-4 i=0$
$\Rightarrow 2 x+i x+2 i y-y+i x-2 x+y+2 i y-4 i=0$
$\Rightarrow 2 i x+4 i y-4 i=0$
$L_{2}: x+2 y-2=0$
Solve $L_{1}$ and $L_{2} \quad 4 y=2, \quad y=\frac{1}{2}$
$\therefore \mathrm{x}=1$
Centre $\left(1, \frac{1}{2}\right)$
$\mathrm{L}_{3}: \mathrm{iz}+\overline{\mathrm{z}}+1+\mathrm{i}=0$
$\Rightarrow \mathrm{i}(\mathrm{x}+\mathrm{i} y)+\mathrm{x}-\mathrm{i} \mathrm{y}+1+\mathrm{i}=0$
$\Rightarrow \mathrm{ix}-\mathrm{y}+\mathrm{x}-\mathrm{i} y+1+\mathrm{i}=0$
$\Rightarrow(\mathrm{x}-\mathrm{y}+1)+\mathrm{i}(\mathrm{x}-\mathrm{y}+1)=0$
Radius $=$ distance from $\left(1, \frac{1}{2}\right)$ to $\mathrm{x}-\mathrm{y}+1=0$
$\mathbf{r}=\frac{1-\frac{1}{2}+1}{\sqrt{2}}$
$\mathrm{r}=\frac{3}{2 \sqrt{2}}$
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