To find the equation of tangent and normal to the circle $x^{2} + y^{2} - 3 x + 4 y - 31 = 0$ at the point $(2, 3)$ .

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Solution

$x^{2} + y^{2} - 3 x + 4 y - 31 = 0 d i f f e r e n t i a t e w . r t o x \Rightarrow 2 x + 2 y d y / d x - 3 + 4 d y / d x = 0 \Rightarrow d y / d x (2 y + 4) + 2 x - 3 = 0 ∴ d y / d x = \frac{3 - 2 x}{(2 y + 4)} \Rightarrow {(d y / d x)}_{2 / 3} = \frac{3 - 4}{6 + 4} = \frac{- 1}{10} \frac{d x}{d y} = - 10 ∴ e q u a t i o n o f tan g e n t = \Rightarrow (y - 3) = d y / d x (x - 2) \Rightarrow y - 3 = \frac{- 1}{10} (x - 2), 10 y - 30 = x + 2, x + 10 y = 32 e q u a t i o n o f n o r m a l \Rightarrow (y - 3) = \frac{d x}{d y} (x - 2), y - 3 = + 10 (x - 2) \Rightarrow 10 x - y = 17$

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