If the locus of , such that , is a circle of radius r and center (a, b), then is equal to: [2025]

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Solution

Q.
If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} - i}) = 2$ , is a circle of radius r and center (a, b), then $\frac{15 a b}{r^{2}}$ is equal to: [2025]

1 12

2 16

3 24

4 18

Ans.
(4)

We have, $Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z} - 1}{2 \bar{z} - i}) = 2$

$\Rightarrow Re (\frac{z - 1}{2 z + i}) + R e (\frac{\bar{z - 1}}{2 z + i}) = 2$ $[∵ (\frac{\bar{z - 1}}{2 z + i}) = \frac{\bar{z} - 1}{2 \bar{z} - i}]$

$\Rightarrow 2 Re (\frac{z - 1}{2 z + i}) = 2 \Rightarrow Re (\frac{z - 1}{2 z + i}) = 1$

Let z = x + iy

$Re (\frac{(x - 1) + i y}{2 x + i (2 y + 1)}) = 1$

$\Rightarrow R e [\frac{((x - 1) + i y) (2 x - i (2 y + 1))}{(2 x + i (2 y + 1)) (2 x - i (2 y + 1))}] = 1$

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

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

$\Rightarrow 2 x^{2} + 2 y^{2} + 3 y + 2 x + 1 = 0$

$\Rightarrow x^{2} + y^{2} + x + \frac{3}{2} y + \frac{1}{2} = 0$

$∴ Centre = (\frac{- 1}{2}, \frac{- 3}{4}) and r = \sqrt{\frac{1}{4} + \frac{9}{16} - \frac{1}{2}} = \frac{\sqrt{5}}{4}$

$\Rightarrow a = \frac{- 1}{2}, b = \frac{- 3}{4}, r^{2} = \frac{5}{16}$

$∴ \frac{15 a b}{r^{2}} = 15 \times (\frac{- 1}{2}) \times (\frac{- 3}{4}) \times \frac{16}{5} = 18$ .

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