Let the lines `(2-i)z=(2+i) barz` and `(2+i)z+(i-2)barz-4i=0`, (here `i^(2)= -1`) be normal to a circle C. If the line `iz+barz+1+i=0` is tangent to this circle C, then its radius is :

Options

A. `(3)/(sqrt(2))`
B. `(1)/(2sqrt(2))`
C. `3sqrt(2)`
D. `(3)/(2sqrt(2))`

Answer

To solve the problem, we need to analyze the given lines and their relationship with the circle.

### Step 1: Understand the given lines
The first line is given by:
\[ (2 - i)z = (2 + i) \bar{z} \]
Rearranging gives:
\[ (2 - i)z - (2 + i) \bar{z} = 0 \]
This can be rewritten as:
\[ (2 - i)z - (2 + i) \left( \frac{x - iy}{1} \right) = 0 \]
where \( z = x + iy \) and \( \bar{z} = x - iy \).