Question

If a complex number \( z \) is such that \( \frac{z-2i}{z-2} \) and the locus of \( z \) is a closed curve, then the area of the region bounded by that closed curve and lying in the first quadrant is:

• A. \( 2\pi \)
• B. \( \frac{\pi}{2} \)
• C. \( \pi \)
• D. \( \frac{\pi}{4} \)

Hint:

The area of the region bounded by a circle in the complex plane can be found by calculating one-fourth of the area of the circle for the first quadrant.

Answer 1

The Correct Option is B

We are given a complex number \( z \) such that: \[ \left| \frac{z - 2i}{z - 2} \right| = 1 \]

 Step 1: Understanding the Given Condition 
Recall that for a complex number, \[ \left| \frac{z - z_1}{z - z_2} \right| = 1 \] This represents the locus of points \( z \) that are equidistant from two fixed points \( z_1 \) and \( z_2 \). In our case, - \( z_1 = 2i \) (point on the imaginary axis) - \( z_2 = 2 \) (point on the real axis) The given equation represents the **perpendicular bisector** of the line segment joining these points, which is a circle passing through \( z_1 \) and \( z_2 \) with its center on the line joining these points. 

Step 2: Determining the Circle's Properties 
The line segment joining \( 2i \) and \( 2 \) has its midpoint at: \[ \left( \frac{2 + 0}{2}, \frac{0 + 2i}{2} \right) = (1, i) \] The radius of the circle is half the distance between these points: \[ \text{Radius} = \frac{\sqrt{(2 - 0)^2 + (0 - 2)^2}}{2} = \frac{\sqrt{4 + 4}}{2} = \frac{\sqrt{8}}{2} = \sqrt{2} \] Thus, the circle has: - Center at \( (1, i) \)
- Radius \( \sqrt{2} \)
Step 3: Finding the Area in the First Quadrant 
Since the circle is symmetric about both axes, the area of the circle is: \[ \text{Total Area} = \pi r^2 = \pi (\sqrt{2})^2 = 2\pi \] Since the first quadrant contains one-fourth of the total area, \[ \text{Area in the first quadrant} = \frac{1}{4} \times 2\pi = \frac{\pi}{2} \]

 Final Answer: (2) \( \frac{\pi}{2} \)