The area (in sq. units) of the part of the ci | vedclass

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(D) The circle is $x^2+y^2=13^2$ and the line is $y=5x-13$. The intersection points are found by substituting $y$ into the circle equation: $x^2+(5x-1

Vedclass · Accepted Answer

$171$

Vedclass · Answer

(D) The circle is $x^2+y^2=13^2$ and the line is $y=5x-13$. The intersection points are found by substituting $y$ into the circle equation: $x^2+(5x-13)^2=169 \implies x^2+25x^2-130x+169=169 \implies 26x^2-130x=0 \implies 26x(x-5)=0$. Thus,$x=0$ (giving $y=-13$) and $x=5$ (giving $y=12$).The area below the line $y=5x-13$ and inside the circle is the area bounded by the circle arc and the line segment. Integrating with respect to $x$ from $0$ to $5$: Area $= \int_{0}^{5} (\sqrt{169-x^2} - (5x-13)) dx$$= \int_{0}^{5} \sqrt{13^2-x^2} dx - \int_{0}^{5} (5x-13) dx$$= [\frac{x}{2}\sqrt{169-x^2} + \frac{169}{2}\sin^{-1}(\frac{x}{13})]_0^5 - [\frac{5x^2}{2}-13x]_0^5$$= (\frac{5}{2}\sqrt{144} + \frac{169}{2}\sin^{-1}(\frac{5}{13})) - (\frac{125}{2}-65)$$= 30 + \frac{169}{2}\sin^{-1}(\frac{5}{13}) + \frac{5}{2} = \frac{65}{2} + \frac{169}{2}\sin^{-1}(\frac{5}{13})$.Using $\sin^{-1}(\frac{5}{13}) = \cos^{-1}(\frac{12}{13}) = \frac{\pi}{2} - \sin^{-1}(\frac{12}{13})$:Area $= \frac{65}{2} + \frac{169}{2}(\frac{\pi}{2} - \sin^{-1}(\frac{12}{13})) = \frac{169\pi}{4} - \frac{169}{2}\sin^{-1}(\frac{12}{13}) + \frac{65}{2}$.Comparing with the form $\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$,we identify $\alpha=169, \beta=2$. Since $\gcd(169, 2)=1$,$\alpha+\beta = 169+2 = 171$.

The area (in sq. units) of the part of the circle $x^2+y^2=169$ which is below the line $5x-y=13$ is $\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$ where $\alpha, \beta$ are coprime numbers. Then $\alpha+\beta$ is equal to . . . . . . .

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