JEE Mains · Maths · STD 12 - 8. Application and integration
The area (in sq. units) of the part of circle \(\mathrm{x}^2+\mathrm{y}^2=169\) which is below the line \(5 \mathrm{x}-\mathrm{y}=13\) 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 ...........
- A \(137\)
- B \(711\)
- C \(271\)
- D \(171\)
Answer & Solution
Correct Answer
(D) \(171\)
Step-by-step Solution
Detailed explanation
\( \text { Area }=\int_{-13}^{12} \sqrt{169-y^2} d y-\frac{1}{2} \times 25 \times 5 \) \( =\frac{\pi}{2} \times \frac{169}{2}-\frac{65}{2}+\frac{169}{2} \sin ^{-1} \frac{12}{13} \) \( \therefore \alpha+\beta=171\)
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