Find the area of the triangle formed by the normal at $(3, - 4)$ to the circle $x^{2} + y^{2} - 22 x - 4 y + 25 = 0$ with the coordinate axes.

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Solution

Normal of a circle at given point $(x_{1}, y_{1}) = y - y_{1} = \frac{y_{1} + f}{x_{1} + g} (x - x_{1})$

here $g = - 11, f = 2$

$\Rightarrow$ equation of normal : $y - (- 4) = \frac{- 4 + 2}{2 - 11} (x - 3)$

$\Rightarrow y + 4 = \frac{2}{9} (x - 3)$

$\Rightarrow 9 y + 36 = 2 x - 6$

$\Rightarrow 2 x - 9 y - 42 = 0$

X-intercept of given normal= $\frac{42}{2} = 21$
Y-intercept of given normal= $\frac{- 42}{9} = - \frac{14}{3}$

$\Rightarrow$ Area= $\frac{1}{2} \times 21 \times \frac{14}{3} = 49$ $c m^{2}$

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