The equation of normal of $x^{2} + y^{2} - 2 x + 4 y - 5 = 0$ at $(2, 1)$ is

y = 3 x - 5

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2 y = 3 x - 4

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y = 3 x + 4

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y = x + 1

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Solution

The correct option is A $y = 3 x - 5$
Given equation is $x^{2} + y^{2} - 2 x + 4 y - 5 = 0$
On differentiating, we get
$2 x + 2 y \frac{d y}{d x} - 2 + 4 \frac{d y}{d x} = 0$
$\Rightarrow (y + 2) \frac{d y}{d x} = 1 - x$
$\Rightarrow {(\begin{matrix} \frac{d y}{d x} \end{matrix})}_{(2, 1)} = \frac{1 - 2}{1 + 2} = \frac{- 1}{3}$
$\Rightarrow - {(\begin{matrix} \frac{d x}{d y} \end{matrix})}_{(2, 1)} = 3$
Now, equation of normal is
$(y - 1) = 3 (x - 2)$
$y - 1 = 3 x - 6$
$\Rightarrow y = 3 x - 5$

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