JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 21)
The area (in sq. units) of the part of the 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 __________.
Answer
171
Explanation
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$$\begin{aligned} & \text { Area }=\int_\limits{-13}^{12} \sqrt{169-y^2} d y-\frac{1}{2} \times 25 \times 5 \\ & =\frac{\pi}{2} \times \frac{169}{2}-\frac{65}{2}+\frac{169}{2} \sin ^{-1} \frac{12}{13} \\ & \therefore \alpha+\beta=171 \end{aligned}$$
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