Question:

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 :

1. (1) $\frac{3}{\sqrt{2}}$

2. (2) $3 \sqrt{2}$

3. (3) $\frac{3}{2 \sqrt{2}}$

4. (4) $\frac{1}{2 \sqrt{2}}$

Correct Option: , 3

Solution:

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

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

$\Rightarrow 2 x-i x+2 i y+y=2 x+i x-2-i y+y$

$\Rightarrow 2 i x-4 i y=0$

$L_{1}: x-2 y=0$

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

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

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

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

$L_{2}: x+2 y-2=0$

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 $\mathrm{x}-\mathrm{y}+1=0$

$\mathbf{r}=\frac{1-\frac{1}{2}+1}{\sqrt{2}}$

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