Question

Find the equation of circle in the complex form which touches the line $iz + \bar z + 1 + i = 0$ and the lines $(2 - i)z = (2 + i)\bar z$ and $(2 + i)z + (i - 2)\bar z - 4i = 0$ are the normal of the circle.
$
  A)|z - (1 + \dfrac{i}{2})| = \dfrac{3}{{\sqrt 2 }} \\
  B)|z - (1 - \dfrac{i}{2})| = \dfrac{3}{2} \\
  C)|z - (1 + \dfrac{i}{2})| = \dfrac{3}{{2\sqrt 2 }} \\
  D)|z - (1 - \dfrac{i}{2})| = \dfrac{3}{{2\sqrt 2 }} \\
 $

Answer 1

Hint: Proceed the solution by finding the centre of the circle by using the equation of normal. Make use of all the given equations according to the requirements.

Complete step-by-step answer:
Given normal of the circle are
$(2 - i)z = (2 + i)\bar z \to (1)$
$(2 + i)z + (i - 2)\bar z - 4i = 0 \to (2)$
From $(1)$ \[\bar z = \dfrac{{(2 - i)z}}{{2 + i}}\]
Let us replace the $\bar z$ value in equation $(2)$ with the help of $(1)$ we get
$\
   \Rightarrow (2 + i)z + \dfrac{{(i - 2)(2 - i)}}{{(2 + i)}}z - 4i = 0 \\
    \\
 $
In the second part of equation let us rationalize with the denominator then we will get
$
   \Rightarrow (2 + i) + \dfrac{{(4i - 3)(2 - i)}}{{(2 + i)(2 - i)}}z - 4i = 0 \\
   \Rightarrow (2 + i) + \dfrac{{(11i - 2)}}{5}z = 4i \\
   \Rightarrow (16i + 8)z = 20i \\
   \Rightarrow z = \dfrac{{5i}}{{4i + 2}} \\
 $
Let us simplify the $z$ value using the denominator we get
$ \Rightarrow \dfrac{{5i(2 - 4i)}}{{(2 + 4i)(2 - 4i)}} = \dfrac{{10i + 20}}{{20}} = 1 + \dfrac{i}{2}$
Therefore the $z$ value =$(1 + \dfrac{i}{2})$
Here z is the centre of the circle. Since the point of intersection of two normal of a circle is the centre of the circle.
Given tangent of circle is $iz + \bar z + 1 + i = 0$
Therefore equation of tangent in standard form
$
   \equiv \dfrac{{iz}}{{(1 + i)}} + \dfrac{{\bar z}}{{(1 + i)}} + 1 = 0 \\
   \equiv (1 + i)z + (1 - i)\bar z + 2 = 0 \\
 $
We know that
Radius of circle=perpendicular distance from centre on tangent
$r = \dfrac{{|\dfrac{{(1 + i)(2 + i)}}{2} + \dfrac{{(1 - i)(2 - i)}}{2} + 2|}}{2}$
On further simplification we get
     Radius (r) = $\dfrac{3}{{2\sqrt 2 }}$
Therefore equation of circle= $|z - (1 + \dfrac{i}{2})| = \dfrac{3}{{2\sqrt 2 }}$
Hence, Option C is the correct answer

Note: Make a note in this problem by using the normal we got the centre of the circle and by the given tangent we got the radius of the circle. Focus on the calculation of the equation as it is of complex numbers