Question

The area (in square units) of the part of the circle $x^2 + y^2 = 169$ which is below the line $5x - y = 13$ is$\frac{\pi \alpha}{2 \beta} - \frac{65}{2} + \frac{\alpha}{\beta} \sin^{-1} \left( \frac{12}{13} \right)$where $\alpha$ and $\beta$ are coprime numbers. Then $\alpha + \beta$ is equal to

Answer 1

Step 1: Identify the Circle and Line Equation

The circle is given by $x^2 + y^2 = 169$, which has a radius of $\sqrt{169} = 13$. The line equation $5x - y = 13$ intersects the circle, creating a segment.

Step 2: Determine Points of Intersection

The line intersects the circle at points $(5, 12)$ and $(0, -13)$, as shown in the solution diagram.

Step 3: Calculate the Area Below the Line

The area of the segment below the line is calculated by integrating from $y = -13$ to $y = 12$:

$\text{Area} = \int_{-13}^{12} \sqrt{169 - y^2} \, dy - \frac{1}{2} \times 25 \times 5$

Step 4: Simplify the Result

After integrating, we get:

$\text{Area} = \frac{\pi}{2} \cdot \frac{169}{2} - \frac{65}{2} + \frac{169}{2} \sin^{-1} \frac{12}{13}$

Step 5: Determine $\alpha$ and $\beta$

Comparing terms, we find $\alpha = 169$ and $\beta = 2$.

Step 6: Calculate $\alpha + \beta$

$\alpha + \beta = 169 + 2 = 171$

So, the correct answer is: 171