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Question

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


Solution

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

here $$g=-11,\,f=2$$

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

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

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

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

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

$$\Rightarrow $$ Area=$$\dfrac{1}{2}\times 21 \times \dfrac{14}{3}=49$$ $$cm^{2}$$

Maths

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