# 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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