Find the equation of the normal to the circle | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 5x + 2y - 48 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (-g, -f) = (2.

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$14x - 5y - 40 = 0$

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(B) The equation of the circle is $x^2 + y^2 - 5x + 2y - 48 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $C = (-g, -f) = (2.5, -1)$. The normal to a circle at any point always passes through the center of the circle. The slope of the normal passing through $(5, 6)$ and $(2.5, -1)$ is given by $m = \frac{6 - (-1)}{5 - 2.5} = \frac{7}{2.5} = \frac{14}{5}$. The equation of the normal is $y - 6 = \frac{14}{5}(x - 5)$. $5(y - 6) = 14(x - 5)$ $5y - 30 = 14x - 70$ $14x - 5y - 40 = 0$.

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

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