Find the equation of the normal to the circle $x^{2} + y^{2} - 5 x + 2 y - 48 = 0$ at the point (5, 6)

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Solution

Since normal is line joining centre $(\frac{5}{2}, - 1)$ and $(5, 6)$ . Slope = $\frac{14}{5}$
Hence the equation of the normal at (5, 6) is $y - 6 = (14 / 5) (x - 5)$ $\Rightarrow$ $14 x - 5 y - 40 = 0$

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