The equation of the normal to the circle $x^{2} + y^{2} - 4 x + 6 y = 0$ at $(0, 0)$ is

3 x - 2 y = 0

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3 x + 2 y = 0

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2 x - 3 y = 0

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2 x + 3 y = 0

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Solution

The correct option is B $3 x + 2 y = 0$
$\begin{matrix} x^{2} + y^{2} - 4 x + 6 y = 0 2 x + 2 y \frac{d y}{d x} - 4 + 6 \frac{d y}{d x} = 0 (2 y + 6) \frac{d y}{d x} = 4 - 2 x \frac{d y}{d x} = \frac{4 - 2 x}{2 y + 6} N o w, {\frac{d y}{d x}}_{(0, 0)} = \frac{4}{6} = \frac{2}{3} s o l p e o f n o r m a l = \frac{- 3}{2} E q u^{n} o f n o r m a l : y = \frac{- 3}{2} x 2 y + 3 x = 0 H e n c e, t h e o p t i o n B i s t h e c o r r e c t a n s w e r . \end{matrix}$

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