JEE Main 2024 (Online) 29th January Morning ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Application of Integrals
Answer$171$
Solution
Line $y=5x-13$ meets the circle at $(0,-13)$ and $(5,12)$. Region below the line is the minor segment, with central angle $\theta$: $\cos\theta=\frac{0\cdot5+(-13)(12)}{169}=-\frac{12}{13}$, so $\theta=\frac\pi2+\sin^{-1}\frac{12}{13}$, $\sin\theta=\frac{5}{13}$. Area $=\frac{169}{2}(\theta-\sin\theta)=\frac{169\pi}{4}+\frac{169}{2}\sin^{-1}\frac{12}{13}-\frac{65}{2}$. Matching $\frac{\pi\alpha}{2\beta}$: $\frac\alpha\beta=\frac{169}{2}$, coprime, so $\alpha=169,\beta=2$ and $\alpha+\beta=171$.