Key Concepts and Formulas

• Complex to Cartesian Conversion: $z = x + iy$ and $\overline z = x - iy$, where $x$ and $y$ are real numbers.
• Normals of a Circle: Normals to a circle pass through the center. Therefore, intersection of two normals yields the center.
• Perpendicular Distance: The distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Step-by-Step Solution

Step 1: Find the Center of the Circle

We are given two normal lines. The intersection of these lines will give us the center of the circle.

First Normal Line (L₁): $(2 - i)z = (2 + i)\overline z$. We substitute $z = x + iy$ and $\overline z = x - iy$ to convert to Cartesian form.

$(2 - i)(x + iy) = (2 + i)(x - iy)$ $2x + 2iy - ix - i^2y = 2x - 2iy + ix - i^2y$ $2x + 2iy - ix + y = 2x - 2iy + ix + y$ $2iy - ix = -2iy + ix$ $4iy - 2ix = 0$ $2y - x = 0$ ${L_1}: x - 2y = 0$

Second Normal Line (L₂): $(2 + i)z + (i - 2)\overline z - 4i = 0$. Substituting $z = x + iy$ and $\overline z = x - iy$:

$(2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0$ $2x + 2iy + ix - y + ix + y - 2x + 2iy - 4i = 0$ $2ix + 4iy - 4i = 0$ $x + 2y - 2 = 0$ ${L_2}: x + 2y - 2 = 0$

Intersection of Normals: We solve the system of equations:

1. $x - 2y = 0 \implies x = 2y$
2. $x + 2y - 2 = 0$

Substituting $x = 2y$ into the second equation: $2y + 2y - 2 = 0$ $4y = 2$ $y = \frac{1}{2}$

Then, $x = 2y = 2\left(\frac{1}{2}\right) = 1$. Thus, the center of the circle is $\left(1, \frac{1}{2}\right)$.

Step 2: Find the Equation of the Tangent Line

The tangent line is given by $iz + \overline z + 1 + i = 0$. Substituting $z = x + iy$ and $\overline z = x - iy$:

$i(x + iy) + (x - iy) + 1 + i = 0$ $ix - y + x - iy + 1 + i = 0$ $(x - y + 1) + i(x - y + 1) = 0$ For this complex number to be zero, both the real and imaginary parts must be zero, leading to the same equation: $x - y + 1 = 0$

Step 3: Calculate the Radius

The radius is the perpendicular distance from the center $\left(1, \frac{1}{2}\right)$ to the tangent line $x - y + 1 = 0$. Using the distance formula:

$r = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} = \frac{\left|1(1) - 1\left(\frac{1}{2}\right) + 1\right|}{\sqrt{1^2 + (-1)^2}}$ $r = \frac{\left|1 - \frac{1}{2} + 1\right|}{\sqrt{2}} = \frac{\left|\frac{3}{2}\right|}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when expanding and simplifying complex expressions and Cartesian equations.
• Real vs. Imaginary: When converting to Cartesian form, correctly separate real and imaginary terms. Setting both real and imaginary parts to zero is crucial.
• Visualizing the Geometry: Understand that the intersection of normals gives the center of the circle, and the radius is the perpendicular distance from the center to the tangent.

Summary

We converted the equations of the normal lines and the tangent line from the complex plane to the Cartesian plane. By finding the intersection of the normal lines, we determined the center of the circle. Then, we calculated the perpendicular distance from the center to the tangent line, which gave us the radius of the circle.

The final answer is \boxed{\frac{3}{2\sqrt{2}}}, which corresponds to option (A).