Question

Let the lines $(2 - i)z = (2 + i)\bar z$ and $(2 + i)z + (i - 2)\bar z - 4i = 0$ be normal to a circle C. If the line $iz + \bar z + 1 + i = 0$ is tangent to this circle C, then its radius is
A. $\dfrac{3}{{\sqrt 2 }}$
B. $3\sqrt 2 $
C. $\dfrac{3}{{2\sqrt 2 }}$
D. $\dfrac{1}{{2\sqrt 2 }}$

Answer 1

Hint: A complex number contains both imaginary and real part. Here, the equations given to us are in terms of the complex numbers, so we put the value of $z$ and $\bar z$ in the equations and then solve to find the centre of the circle C and then we will find the radius of the circle using the distance formula.

Formula used: The distance of a point $({x_0},{y_0})$ from a line $Ax + By + C = 0$ is $d = \dfrac{{\left| {A{x_0} + B{y_0} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$

Complete step by step answer:
We are given the equations of two lines as follows:
$
  (2 - i)z = (2 + i)\bar z \\
  (2 + i)z + (i - 2)\bar z - 4i = 0 \\
 $
We know that $z = x + iy$ is a complex number and its conjugate is $\bar z = x - iy$. Putting these two values in the first equation, we get:
$
  (2 - i)(x + iy) = (2 + i)(x - iy) \\
   \Rightarrow 2x + 2iy - ix + y = 2x - 2iy + ix + y\,\,\,\,\,\,(\because {i^2} = - 1) \\
   \Rightarrow 4iy - 2ix = 0 \\
   \Rightarrow x - 2y = 0 \\
 $
Now, putting the values in the second equation, we get:
$
  (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0 \\
   \Rightarrow 2x + 2iy + ix - y + ix + y - 2x + 2iy - 4i = 0 \\
   \Rightarrow 4iy + 2ix - 4i = 0 \\
   \Rightarrow x + 2y - 2 = 0 \\
 $
Adding the two obtained equations, we get:
$
  x - 2y + x + 2y - 2 = 0 \\
   \Rightarrow 2x = 2 \\
   \Rightarrow x = 1 \\
 $
Putting the value of $x$ in the first equation, we get $y = \dfrac{1}{2}$
As we know that the given two lines are normal to a circle, so they will intersect at its centre and thus the centre is $(1,\dfrac{1}{2})$.
Now, line $iz + \bar z + 1 + i = 0$ is tangent to the given circle. Simplifying this equation, we get:
$
  i(x + iy) + x - iy + 1 + i = 0 \\
   \Rightarrow ix - y + x - iy + 1 + i = 0 \\
   \Rightarrow i(x - y + 1) + x - y + 1 = 0 \\
   \Rightarrow x - y + 1 = 0 \\
 $
As this line is tangent to the circle so the distance between the point $(1,\dfrac{1}{2})$ and the line $x - y + 1 = 0$ will be the radius of the circle.
We know that distance of a point $({x_0},{y_0})$ from a line $Ax + By + C = 0$ is $d = \dfrac{{\left| {A{x_0} + B{y_0} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$
So, radius:
$
  R = \dfrac{{\left| {1(1) - (1)(\dfrac{1}{2}) + 1} \right|}}{{\sqrt {{1^2} + {{( - 1)}^2}} }} \\
   \Rightarrow R = \dfrac{3}{{2\sqrt 2 }} \\
 $
The correct option is option C.

Note: While putting the value of $z$ and $\bar z$ , note that $z = x + iy$ and $\bar z$ is its conjugate, that is, $\bar z = x - iy$ .