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:

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Answer: B. 100 \begin{aligned} & \left|\frac{\bar{z}-i}{2 \bar{z}+i}ight|=\frac{1}{3} \ & \left|\frac{\bar{z}-i}{\bar{z}+\frac{i}{2}}ight|=\frac{2}{3} \ & 3|x-i y-i|=2\left|x-i y+\frac{i}{2}ight| \ & 9\left(x^2+(y+1)^2ight)=4\left(x^2+(y-1 / 3)^2ight) \ & 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 \ & 	ext { centre } \Rightarrow\left(0,-\frac{11}{5}ight) \end{aligned} \left| {{1 \over 2}\left| {\begin{matrix} 0 & 0 & 1 \ 0 & { - 11/5} & 1 \ \alpha & 0 & 1 \ \end{matrix} } ight|} ight| = 11 \begin{aligned} & \Rightarrow\left(-\frac{11}{5} \alphaight)^2=(11 	imes 2)^2 \ & \Rightarrow \alpha^2=100 \end{aligned}

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:

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