Area Under Curves Question 6
Question 6 - 2024 (29 Jan Shift 1)
The area (in sq. units) of the part of circle $x^{2}+y^{2}=169$ which is below the line $5 x-y=13$ 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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Answer (171)
Solution
Area $=\int_{-13}^{12} \sqrt{169-y^{2}} d y-\frac{1}{2} \times 25 \times 5$ $=\frac{\pi}{2} \times \frac{169}{2}-\frac{65}{2}+\frac{169}{2} \sin ^{-1} \frac{12}{13}$ $\therefore \alpha+\beta=171$
