The area (in sq. units) of the part of circle x ^ 2 + y ^ 2 = 169...

Given: ${x}^{2} + {y}^{2} = 169, 5x - y = 13$

$⇒ {x}^{2} + {\left(5x - 13\right)}^{2} = 169$

$⇒ {x}^{2} + 25{x}^{2} + 169 - 130x - 169 = 0$

$⇒ 26{x}^{2} - 130x = 0$

$⇒ x\left(x - 5\right) = 0$

$⇒ x = 0, 5$

$⇒ y = - 13, 12$

So, the points of intersection are \left(0, - 13\right)\text{and}\left(5, 12\right)$.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/7c2a25f8-2af4-4c12-890f-d36aefb3bfe0.png" style="width: 375px; height: 258px;"/></p><p>So, the required area is given by,</p><p>$A = {∫}_{- 13}^{12}\sqrt{169 - {y}^{2}}dy - \dfrac{1}{2} \times 25 \times 5$</p><p>$⇒ A = {\left[\dfrac{x}{2}\sqrt{169 - {x}^{2}} + \dfrac{169}{2}\sin^{- 1}\dfrac{x}{13}\right]}_{- 13}^{12} - \dfrac{1}{2} \times 25 \times 5$</p><p>$⇒ A = \dfrac{\pi}{2} \times \dfrac{169}{2} - \dfrac{65}{2} + \dfrac{169}{2}\sin^{- 1}\dfrac{12}{13}$</p><p>Hence, on comparing with given value we get,</p><p>$⇒ \alpha = 169, \beta = 2$</p><p>$⇒ \alpha + \beta = 171$