Let $$\mid\frac{\overline{z}-i}{2\overline{z}+i}\mid=\frac{1}{3}, z\in C$$, 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 $$(\alpha,0)$$ is 11 square units, then $$\alpha^{2}$$ equals:

Given $$\left|\frac{\bar{z}-i}{2\bar{z}+i}\right| = \frac{1}{3}$$, find $$\alpha^2$$ where the area of the triangle with vertices $$(0,0)$$, center $$C$$, and $$(\alpha, 0)$$ is 11.

Let $$z = x + iy$$, so $$\bar{z} = x - iy$$.

$$\left|\frac{(x-iy) - i}{2(x-iy) + i}\right| = \frac{1}{3}$$

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

Squaring and using $$|w_1/w_2|^2 = |w_1|^2/|w_2|^2$$:

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

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

$$9x^2 + 9y^2 + 18y + 9 = 4x^2 + 4y^2 - 4y + 1$$

$$5x^2 + 5y^2 + 22y + 8 = 0$$

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

Completing the square: $$x^2 + \left(y + \frac{11}{5}\right)^2 = \frac{121}{25} - \frac{8}{5} = \frac{121 - 40}{25} = \frac{81}{25}$$

Center $$C = \left(0, -\frac{11}{5}\right)$$, radius $$= \frac{9}{5}$$.

Vertices: $$(0,0)$$, $$\left(0, -\frac{11}{5}\right)$$, $$(\alpha, 0)$$.

Area = $$\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

$$= \frac{1}{2}\left|0 + 0 + \alpha\left(0 + \frac{11}{5}\right)\right| = \frac{11|\alpha|}{10}$$

Setting this equal to 11: $$\frac{11|\alpha|}{10} = 11 \implies |\alpha| = 10$$

$$\alpha^2 = 100$$

The correct answer is Option 2: 100.