Question

Let the lines \( (2 - i)z = (2 + i)\bar{z} \) and \( (2 + i)z + (i - 2)\bar{z} - 4i = 0 \), (here \( i^2 = -1 \)) 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. 3/√2
• B. 1/√2
• C. 3/2√2
• D. 3/√2

Hint:

The center of a circle is the point of intersection of any two of its normals.

Answer 1

The Correct Option is C

Step 1: Find the center of the circle by intersecting the normals. Line 1: $(2-i)(x+iy) = (2+i)(x-iy) \Rightarrow 2x+2iy-ix+y = 2x-2iy+ix+y \Rightarrow 4iy = 2ix \Rightarrow x=2y$. Line 2: $(2+i)(x+iy) + (i-2)(x-iy) = 4i$. Simplifying gives $x+2y=2$.
Step 2: Solving $x=2y$ and $x+2y=2$ gives $y=1/2, x=1$. Center is $(1, 1/2)$.
Step 3: Convert tangent $iz + \bar{z} + 1 + i = 0$ to Cartesian: $i(x+iy) + (x-iy) + 1 + i = 0 \Rightarrow (x-y+1) + i(x-y+1) = 0 \Rightarrow x-y+1=0$.
Step 4: Radius $r = \frac{|1 - 1/2 + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{3/2}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$.