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 π α/2 β-65/2+α/β sin -1 (12/13 ) where α, β are coprime numbers. Then α+β is equal to

Mathematics · Area under the Curves

JEE Main 2024 — 29 January, Shift 1 — Question 27

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 $\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right) \quad$ where $\alpha, \beta \quad$ are coprime numbers. Then $\alpha+\beta$ is equal to

Answer: 171

Numerical answer — enter this value.

Step-by-step solution

Area $=\int_{-13}^{12} \sqrt{169-y^{2}} d y-\frac{1}{2} \times 25 \times 5$

$=\frac{\pi}{2} \times \frac{169}{2}-\frac{65}{2}+\frac{169}{2} \sin ^{-1} \frac{12}{13}$

$\therefore \alpha+\beta=171$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves