Question

Let the lines $(2–i)z=(2+i)$ and $(2+i)z+(i–2)–4i=0$, (here $i^2=–1$) be normal to a circle $C$. If the line$iz+1+i=0$ 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

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

Explanation for the correct answer:

Determine the radius of the circle:

Given,

$$\begin{gathered} \Rightarrow (2–i)z=(2+i) \\ \Rightarrow (2–i)(x+iy)=(2+i)(x–iy) \\ \Rightarrow 2x–ix+2iy+y=2x+ix–2iy+y \\ \Rightarrow 2ix–4iy=0 \\ \Rightarrow x-2y=0\left[\mathrm{Divide}\mathrm{by}2i\right] \\ L1:x–2y=0 \end{gathered}$$

We also have been given,

$$\begin{gathered} \Rightarrow (2+i)z+(i–2)–4i=0 \\ \Rightarrow (2+i)(x+iy)+(i–2)(x–iy)–4i=0 \\ \Rightarrow 2x+ix+2iy–y+ix–2x+y+2iy–4i=0 \\ \Rightarrow 2ix+4iy–4i=0 \\ \Rightarrow x+2y-2=0\left[\mathrm{Divide}\mathrm{by}2i\right] \\ L2:x+2y–2=0 \end{gathered}$$

Adding L₁ and L₂

$$\begin{gathered} \Rightarrow -2y+x+2y-2=0 \\ \Rightarrow 2x-2=0 \\ \Rightarrow 2x=2 \\ \Rightarrow x=1 \\ \mathrm{now}, \\ \Rightarrow x-2y=0 \\ \Rightarrow 1-2y=0 \\ \Rightarrow 2y=1 \\ \Rightarrow y=\frac{1}{2} \end{gathered}$$

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

Given that,

$$\begin{gathered} \Rightarrow L_3:iz+1+i=0 \\ \Rightarrow i(x+iy)+x–iy+1+i=0 \\ \Rightarrow ix–y+x–iy+1+i=0 \\ \Rightarrow (x–y+1)+i(x–y+1)=0 \\ \mathrm{We}\mathrm{get} \\ \mathrm{x}+\mathrm{y}-1=0 \end{gathered}$$

So the radius = distance from $\left(1,\frac{1}{2}\right)$ to $x–y+1=0$

Using formula $\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}$, where $\left(x_1,y_1\right)=\left(1,\frac{1}{2}\right)\mathrm{and}Ax+By+C=x-y+1$

$$\begin{gathered} r=\frac{1-{\displaystyle \frac{1}{2}}+1}{\sqrt{2}} \\ r=\frac{3}{2\sqrt{2}} \end{gathered}$$

Therefore, the correct answer is option (C).