JEE Main 2025 — 23 January, Morning Shift — Question 10

Let $z = x + iy \implies \bar{z} = x - iy$. The given equation is

$$\left|\frac{\bar{z}-i}{2\bar{z}+i}\right| = \frac{1}{3} \implies \left|\frac{x - i(y+1)}{2x - i(2y-1)}\right| = \frac{1}{3}.$$

Taking modulus and squaring,

$$\frac{x^2 + (y+1)^2}{4x^2 + (2y-1)^2} = \frac{1}{9} \implies 9(x^2 + (y+1)^2) = 4x^2 + (2y-1)^2.$$

Simplifying,

$$5x^2 + 5y^2 + 22y + 8 = 0 \implies x^2 + y^2 + \frac{22}{5}y + \frac{8}{5} = 0.$$

Completing the square:

$$x^2 + \left(y + \frac{11}{5}\right)^2 = \frac{81}{25}.$$

So the circle has center $C = \left(0, -\frac{11}{5}\right)$. The area of the triangle with vertices $(0,0), C, (\alpha,0)$ is

$$\text{Area} = \frac{1}{2} \left| \alpha \left(0 - \left(-\frac{11}{5}\right)\right) \right| = \frac{11}{10} |\alpha|.$$

Setting area $= 11$ gives $|\alpha| = 10 \implies \alpha^2 = 100$.