Find the equation of normal to the circle $x^{2} + y^{2} - 5 x + 2 y - 48 = 0$ which is parallel to the line 14x - 5y - 30 = 0

14x - 5y + 20 = 0

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14x - 5y = 20

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14x - 5y + 40 = 0

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14x - 5y = 40

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Solution

The correct option is D
14x - 5y = 40

We saw that every normal passes through centre. We want to find the equation of normal parallel to 14x-5y-30=0. We know that any line parallel to 14x-5y-30=0 can be written as 14x-5y+k=0.

This passes through center $(\frac{5}{2}, - 1)$

$\Rightarrow 14 \times \frac{5}{2} - 5 (- 1) + k = 0$

$\Rightarrow k = - 40$

$\Rightarrow 14 x - 5 y - 40 = 0$ is the equation of normal

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Similar questions

Find the equation of normal to the circle $x^{2} + y^{2} - 5 x + 2 y - 48 = 0$ which is parallel to the line 14x - 5y - 30 = 0

A (10, 4), B (- 4, 9)

C (- 2, - 1)

x^{2} + y^{2} - 6 x - 4 y + 4 = 0

x^{2} + y^{2} - 14 x - 5 y - 24 = 0

L_{1} : \frac{x - 1}{2} = \frac{y}{- 2} = \frac{z + 2}{- 1}

L_{2} : \frac{x + 3}{8} = \frac{y - 4}{1} = \frac{z}{- 4}

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