Let , be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the point (0, 0), C and is 11 square units, then equals : [2025]

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Solution

Q.
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 point (0, 0), C and $(α, 0)$ is 11 square units, then $α^{2}$ equals : [2025]

1 100

2 $\frac{81}{25}$

3 $\frac{121}{25}$

4 50

Ans.
(1)

Given, $| \frac{\bar{z} - i}{2 \bar{z} + i} |$ $= \frac{1}{3} and z \in C$

$\Rightarrow$ $| \frac{\bar{z} - i}{\bar{z} + \frac{i}{2}} |$ $= \frac{2}{3} \Rightarrow 3$ $| x - i y - i |$ $= 2$ $| x - i y + \frac{i}{2} |$

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

$\Rightarrow 3 \sqrt{x^{2} + {(y + 1)}^{2}} = 2 \sqrt{{(x)}^{2} + {(y - \frac{1}{2})}^{2}}$

Squaring on both sides, we get

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

$\Rightarrow 9 x^{2} + 9 y^{2} + 9 + 18 y = 4 x^{2} + 4 y^{2} + 1 - 4 y$

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

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

Centre $(C) = (0, - \frac{11}{5})$

Now, $\frac{1}{2}$ $| 0 + 0 + α (0 + \frac{11}{5}) |$ $= 11$ [Given]

$\Rightarrow \frac{11 α}{5} = 22 \Rightarrow α = 10$ $∴ α^{2} = 100$ .

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