[UPLIFTJEE]

Let $\left|\frac{\bar{z}-\mathrm{i}}{2 \bar{z}+\mathrm{i}}\right|=\frac{1}{3}, \mathrm{z} \in \mathrm{C}$, 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 [JEE(Main) 23 Jan 2025 S I]
(A) 100
(B) $\frac{81}{25}$
(C) $\frac{121}{25}$
(D) 50

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Solution

$\left|\frac{\bar{z}-\mathrm{i}}{2 \bar{z}+\mathrm{i}}\right|=\frac{1}{3} \Rightarrow\left|\frac{\bar{z}-\mathrm{i}}{\bar{z}+\frac{\mathrm{i}}{2}}\right|=\frac{2}{3} \Rightarrow 3|\mathrm{x}-\mathrm{iy}-\mathrm{i}|=2\left|\mathrm{x}-\mathrm{iy}+\frac{\mathrm{i}}{2}\right|$

\[ \Rightarrow 9\left(x^{2}+(y+1)^{2}\right)=4\left(x^{2}+(y-1 / 2)^{2}\right) \Rightarrow 9 x^{2}+9 y^{2}+18 y+9=4 x^{2}+4 y^{2}-4 y+1 \Rightarrow 5 x^{2}+5 y^{2}+22 y+8=0 \]

\[ \begin{aligned} & \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+\frac{22}{5} \mathrm{y}+\frac{8}{5}=0 \text { centre } \mathrm{C}\left(0,-\frac{11}{5}\right) \text { Area of triangle formed by }(0,0),\left(0,-\frac{11}{5}\right) \text { and }(\alpha, 0) \text { is }\left|\frac{1}{2}\right| \begin{array}{ccc} 0 & 0 & 1 \\ 0 & -11 / 5 & 1 \\ \alpha & 0 & 1 \end{array}|\mid=11 \\ & \Rightarrow\left(\frac{11}{5} \alpha\right)^{2}=(11 \times 2)^{2} \Rightarrow \alpha^{2}=100 \end{aligned} \]

[UPLIFTJEE]

Answer

A

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