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Area Under The Curves

JEE Main 2024 (Online) 29th January Morning Shift

INTEGER+4 / -12024

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 __________.

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