JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Complex Numbers

Let

\frac{z ˉ - i}{2 z ˉ + i} = \frac{1}{3}, z \in C

, be the equation of a circle with center at C . If the area of the triangle, whose vertices are at the points

(0, 0), C

and

(α, 0)

is 11 square units, then

α^{2}

equals

Answer: (A)

Solution

Write

z = x + i y

, so

\overset{z}{ˉ} = x - i y

. The condition

\frac{z ˉ - i}{2 z ˉ + i} = \frac{1}{3}

gives

3∣ x - i (y + 1) ∣ = ∣2 x - i (2 y - 1) ∣

Squaring,

9 [x^{2} + (y + 1)^{2}] = 4 x^{2} + (2 y - 1)^{2}

9 x^{2} + 9 y^{2} + 18 y + 9 = 4 x^{2} + 4 y^{2} - 4 y + 1

5 x^{2} + 5 y^{2} + 22 y + 8 = 0 ⟹ x^{2} + y^{2} + \frac{22}{5} y + \frac{8}{5} = 0

This circle has centre

C = (0, - \frac{11}{5})

.

The triangle with vertices

(0, 0)

C

, and

(α, 0)

has base

O C = \frac{11}{5}

along the y-axis and height

∣ α ∣

Area = \frac{1}{2} \cdot \frac{11}{5} \cdot ∣ α ∣ = 11 ⟹ ∣ α ∣ = 10 ⟹ α^{2} = 100

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option