Let the line (2 - i )z = (2 + i)


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 $C$. If the line $i z+\bar{z}+1+i=0$ is tangent to this circle $\mathrm{C}$, then its radius is:

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

  2. $\frac{1}{2 \sqrt{2}}$

  3. $3 \sqrt{2}$

  4. $\frac{3}{2 \sqrt{2}}$

Correct Option: , 4


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


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

$x+2 y=2$

(iii) $\mathrm{iz}+\overline{\mathrm{Z}}+1+\mathrm{i}=0$

Eq $^{n}$ of tangent $x-y+1=0$

Solving (i) and (ii)

$x=1, y=\frac{1}{2}$

Now, $\mathrm{p}=\mathrm{r} \Rightarrow\left|\frac{1-\frac{1}{2}+1}{\sqrt{2}}\right|=\mathrm{r}$

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

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