Question

Let the lines $(2–i)z=(2+i)\overline{z}$ and $(2+i)z+(i–2)\overline{z}–4i=0,$ $($here $i^2=–1$$)$ be normal to a circle $C$. If the line $iz+\overline{z}+1+i=0$ is tangent to this circle $C$, then its radius is :

• A. $\frac{3}{\sqrt{2}}$
• B. $3\sqrt{2}$
• C. $\frac{3}{2\sqrt{2}}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is C $\frac{3}{2\sqrt{2}}$

$(2–i)z=(2+i)\overline{z}\Rightarrow (2–i)(x+iy)=(2+i)(x–iy)\Rightarrow 2x–ix+2iy+y=2x+ix–2iy+y\Rightarrow 2ix–4iy=0L_1:x–2y=0\Rightarrow (2+i)z+(i–2)\overline{z}–4i=0.\Rightarrow (2+i)(x+iy)+(i–2)(x–iy)–4i=0.\Rightarrow 2x+ix+2iy–y+ix–2x+y+2iy–4i=0\Rightarrow 2ix+4iy–4i=0L_2:x+2y–2=0$
Solve $L_1$ and $L_2,4y=2,y=\frac{1}{2},\therefore x=1$
Centre $\equiv (1,\frac{1}{2})$
$L_3:iz+\overline{z}+1+i=0$
$\Rightarrow i(x+iy)+x–iy+1+i=0\Rightarrow ix–y+x–iy+1+i=0\Rightarrow (x–y+1)+i(x–y+1)=0$
Radius $=$ distance from $(1,\frac{1}{2})$ to $x–y+1=0r=\frac{1-\frac{1}{2}+1}{\sqrt{2}}r=\frac{3}{2\sqrt{2}}$