Find the equation of the normal to the circle \(x^2 + y^2 – 5x + 2y -48\) = 0 at the point (5,6).

Solution :

Since the normal to the circle always passes through the center,

so the equation of the normal will be the line passing through (5,6) & (\(5\over 2\), -1)

i.e.  y + 1 = \(7\over {5/2}\)(x – \(5\over 2\)) \(\implies\) 5y + 5 = 14x – 35

\(\implies\)  14x – 5y – 40 = 0

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