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, where are coprime numbers. Then is equal to [2024]

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Solution

Q.
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 $\frac{π α}{2 β} - \frac{65}{2} + \frac{α}{β} \sin^{- 1} (\frac{12}{13}),$ where $α, β$ are coprime numbers. Then $α + β$ is equal to _______ . [2024]

Ans.
(171)

$∴ Required Area = \int_{- 13}^{12} (\sqrt{169 - y^{2}} - \frac{y + 13}{5}) d y$

$= {[\frac{1}{2} y \sqrt{169 - y^{2}} + \frac{169}{2} \sin^{- 1} \frac{y}{13}]}_{- 13}^{12} - {[\frac{y^{2}}{10} + \frac{13}{5} y]}_{- 13}^{12}$

$= [\frac{1}{2} \times 12 \times 5 + \frac{169}{2} \sin^{- 1} \frac{12}{13}] - [0 + \frac{169}{2} (- \frac{π}{2})] - [\frac{144}{10} + \frac{13}{5} \times 12] + [\frac{169}{10} - \frac{169}{5}]$

$= \frac{169}{2} (\frac{π}{2}) + \frac{169}{2} \sin^{- 1} (\frac{12}{13}) - \frac{65}{2}$

$∴$ $α = 169 and β = 2$

$Thus, α + β = 169 + 2 = 171$

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