The area (in sq. units) of the part of 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
Step-by-Step Solution
Key Concept: The circle has radius 13, centre at origin. Line $5x-y=13$ intersects circle at $(5,12)$ and $(0,-13)$. Area below the line = area of circular segment. Use $\int_{-13}^{12}\sqrt{169-y^2}\,dy$ minus the triangle area.
Area $=\int_{-13}^{12}\sqrt{169-y^2}\,dy-\frac{1}{2}\times25\times5=\frac{\pi\times169}{2\times2}-\frac{65}{2}+\frac{169}{2}\sin^{-1}\frac{12}{13}$.
So $\alpha=169$, $\beta=2$, $\alpha+\beta=171$.
Correct Answer: 171