The provided circle equation is \(x^2 + y^2 = 169\), yielding a radius of \(\sqrt{169} = 13\). The line equation is \(5x - y = 13\). These define a segment of the circle.
The line intersects the circle at the points \((5, 12)\) and \((0, -13)\). These are illustrated in the accompanying diagram.
The area of the segment situated below the line is computed via integration from \(y = -13\) to \(y = 12\):
\[ \text{Area} = \int_{-13}^{12} \sqrt{169 - y^2} \, dy - \frac{1}{2} \times 25 \times 5 \]
The integrated result is:
\[ \text{Area} = \frac{\pi}{2} \cdot \frac{169}{2} - \frac{65}{2} + \frac{169}{2} \sin^{-1} \frac{12}{13} \]
By comparing the simplified area expression, we identify \(\alpha = 169\) and \(\beta = 2\).
\[ \alpha + \beta = 169 + 2 = 171 \]
The final answer is: 171
In a △ABC, suppose y = x is the equation of the bisector of the angle B and the equation of the side AC is 2x−y = 2. If 2AB = BC and the points A and B are respectively (4, 6) and (α, β), then α + 2β is equal to: