Let $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}, z \in C$, 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:

<p>$$\begin{aligned} & \left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3} \\ & \left|\frac{\bar{z}-i}{\bar{z}+\frac{i}{2}}\right|=\frac{2}{3} \\ & 3|x-i y-i|=2\left|x-i y+\frac{i}{2}\right| \\ & 9\left(x^2+(y+1)^2\right)=4\left(x^2+(y-1 / 3)^2\right) \\ & 9 x^2+9 y^2+18 y+9=4 x^2+4 y^2-4 y+1 \\ & 5 x^2+5 y^2+22 y+8=0 \\ & x^2+y^2+\frac{22}{5} y+\frac{8}{5}=0 \\ & \text { centre } \Rightarrow\left(0,-\frac{11}{5}\right) \end{aligned}$$</p> <p>$$\left| {{1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr 0 & { - 11/5} & 1 \cr \alpha & 0 & 1 \cr } } \right|} \right| = 11$$</p> <p>$$\begin{aligned} & \Rightarrow\left(-\frac{11}{5} \alpha\right)^2=(11 \times 2)^2 \\ & \Rightarrow \alpha^2=100 \end{aligned}$$</p>