The area (in sq.

Question

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 $$\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$$, where $$\alpha, \beta$$ are cop...

Answer: 171

Accepted Answer

$\begin{gathered} =\int _{-13}^{12}\sqrt{169-{y}^{2}}dy-\frac{1}{2}	imes 25	imes 5 \ =\frac{\pi }{2}	imes \frac{169}{2}-\frac{65}{2}+\frac{169}{2}{\sin }^{-1}\frac{12}{13} \ 	herefore \alpha +\beta =171 \end{gathered}$

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