Let $|\frac{\bar{z} - i}{2 \bar{z} + i}| = \frac{1}{3}, z \in ℂ,$ 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 ( $α$ , 0) is 11 square units, then $α$ ² equals

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100

$\frac{81}{25}$

$\frac{121}{25}$

answer is A.

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Detailed Solution

$\begin{array}{l} |\frac{\bar{z} - i}{2 \bar{z} + i}| = \frac{1}{3} \\ |\frac{\bar{z} - i}{\bar{z} + \frac{i}{2}}| = \frac{2}{3} \\ 3 | x - i y - i | = 2 |x - i y + \frac{i}{2}| \\ 9 (x^{2} + (y + 1)^{2}) = 4 (x^{2} + (y - 1 / 3)^{2}) \\ 9 x^{2} + 9 y^{2} + 18 y + 9 = 4 x^{2} + 4 y^{2} - 4 y + 1 \\ 5 x^{2} + 5 y^{2} + 22 y + 8 = 0 \\ x^{2} + y^{2} + \frac{22}{5} y + \frac{8}{5} = 0 \end{array}$

centre $(0, \frac{- 11}{5})$

$\begin{array}{l} |\frac{1}{2}| \begin{array}{ccc} 0 & 0 & 1 \\ 0 & - 11 / 5 & 1 \\ α & 0 & 1 \end{array}| = 11 \\ \Rightarrow {(- \frac{11}{5} α)}^{2} = (11 \times 2)^{2} \\ \Rightarrow α^{2} = 100 \end{array}$