Consider a circle C 1 : x 2 + y 2 - 4 x - 2 y = - 5 . Let its mirror image in the line y = 2 x + 1 be another circle C 2 : 5 x 2 + 5 y 2 - 10 f x - 10 g y + 36 = 0 . Let r be the radius of C 2 . Then is equal to _______ . [2023]

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Solution

Q.
Consider a circle $C_{1} : x^{2} + y^{2} - 4 x - 2 y = α - 5$ . Let its mirror image in the line $y = 2 x + 1$ be another circle $C_{2} : 5 x^{2} + 5 y^{2} - 10 f x - 10 g y + 36 = 0$ . Let $r$ be the radius of $C_{2}$ . Then $α + r$ is equal to _______ . [2023]

Ans.
(2)

We have,

$C_{1} : x^{2} + y^{2} - 4 x - 2 y - (α - 5) = 0$

$∴$ $Centre is (2, 1), r = \sqrt{4 + 1 + α - 5} = \sqrt{α}$

$C_{2} : 5 x^{2} + 5 y^{2} - 10 f x - 10 g y + 36 = 0$

$i.e., x^{2} + y^{2} - 2 f x - 2 g y + \frac{36}{5} = 0$

$Centre is (f, g), r = \sqrt{f^{2} + g^{2} - \frac{36}{5}}$

$The image of (2, 1) with respect to 2 x - y + 1 = 0 is (f, g),$

$then \frac{f - 2}{2} = \frac{g - 1}{- 1} = \frac{- 2 (4 - 1 + 1)}{5}$

$\Rightarrow \frac{f - 2}{2} = \frac{- 8}{5} and g - 1 = \frac{8}{5}$

$\Rightarrow f - 2 = - \frac{16}{5} and g - 1 = \frac{8}{5}$

$\Rightarrow f = 2 - \frac{16}{5} = \frac{- 6}{5}$ and $g = \frac{13}{5}$

$So, (f, g) = (- \frac{6}{5}, \frac{13}{5})$

$and r = \sqrt{\frac{36}{25} + \frac{169}{25} - \frac{36}{5}} \Rightarrow r = 1$

$\Rightarrow r = 1$

$∴$ $α + r = 1 + 1 = 2$

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