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

The circle is given by $$x^2 + y^2 = 169$$, which has center $$(0,0)$$ and radius $$13$$. The line is $$5x - y = 13$$. The area below this line and inside the circle needs to be found.

First, find the points of intersection between the circle and the line by substituting $$y = 5x - 13$$ into the circle equation:

$$x^2 + (5x - 13)^2 = 169$$

$$x^2 + 25x^2 - 130x + 169 = 169$$

$$26x^2 - 130x = 0$$

$$2x(13x - 65) = 0$$

So, $$x = 0$$ or $$x = 5$$.

When $$x = 0$$, $$y = 5(0) - 13 = -13$$.

When $$x = 5$$, $$y = 5(5) - 13 = 12$$.

The points of intersection are $$(0, -13)$$ and $$(5, 12)$$.

The chord joining these points subtends an angle $$\theta$$ at the center. The vectors are $$\overrightarrow{OA} = (0, -13)$$ and $$\overrightarrow{OB} = (5, 12)$$. The dot product is:

$$\overrightarrow{OA} \cdot \overrightarrow{OB} = (0)(5) + (-13)(12) = -156$$

$$|\overrightarrow{OA}| = 13, \quad |\overrightarrow{OB}| = 13$$

$$\cos \theta = \frac{-156}{13 \times 13} = -\frac{12}{13}$$

Thus, $$\theta = \cos^{-1}\left(-\frac{12}{13}\right)$$. Since $$\cos \theta = -\frac{12}{13}$$, $$\theta$$ is obtuse, and $$\sin \theta = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \sqrt{\frac{25}{169}} = \frac{5}{13}$$ (positive as $$\theta$$ is in the second quadrant).

Let $$\gamma = \cos^{-1}\left(\frac{12}{13}\right)$$, so $$\theta = \pi - \gamma$$ and $$\sin \gamma = \frac{5}{13}$$.

The area of the minor segment (below the chord) is given by the formula $$\frac{r^2}{2} (\theta - \sin \theta)$$:

$$\text{Area} = \frac{169}{2} \left( \theta - \frac{5}{13} \right)$$

Substitute $$\theta = \pi - \gamma$$:

$$\text{Area} = \frac{169}{2} \left( (\pi - \gamma) - \frac{5}{13} \right) = \frac{169}{2} (\pi - \gamma) - \frac{169}{2} \cdot \frac{5}{13}$$

$$\frac{169}{2} \cdot \frac{5}{13} = \frac{169 \times 5}{2 \times 13} = \frac{13 \times 13 \times 5}{2 \times 13} = \frac{65}{2}$$

So,

$$\text{Area} = \frac{169}{2} (\pi - \gamma) - \frac{65}{2}$$

Now, $$\gamma = \cos^{-1}\left(\frac{12}{13}\right) = \sin^{-1}\left(\frac{5}{13}\right)$$. Let $$\beta = \sin^{-1}\left(\frac{12}{13}\right)$$, so $$\sin \beta = \frac{12}{13}$$ and $$\cos \beta = \frac{5}{13}$$. Then $$\gamma = \frac{\pi}{2} - \beta$$, since $$\sin(\gamma + \beta) = \sin \gamma \cos \beta + \cos \gamma \sin \beta = \left(\frac{5}{13}\right)\left(\frac{5}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{12}{13}\right) = 1$$, and $$\gamma + \beta = \frac{\pi}{2}$$.

Substitute $$\gamma = \frac{\pi}{2} - \beta$$:

$$\text{Area} = \frac{169}{2} \left( \pi - \left(\frac{\pi}{2} - \beta\right) \right) - \frac{65}{2} = \frac{169}{2} \left( \pi - \frac{\pi}{2} + \beta \right) - \frac{65}{2}$$

$$= \frac{169}{2} \left( \frac{\pi}{2} + \beta \right) - \frac{65}{2} = \frac{169}{2} \cdot \frac{\pi}{2} + \frac{169}{2} \beta - \frac{65}{2} = \frac{169\pi}{4} + \frac{169}{2} \beta - \frac{65}{2}$$

Since $$\beta = \sin^{-1}\left(\frac{12}{13}\right)$$,

$$\text{Area} = \frac{169\pi}{4} + \frac{169}{2} \sin^{-1}\left(\frac{12}{13}\right) - \frac{65}{2}$$

The given expression 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 integers. Comparing coefficients:

Coefficient of $$\pi$$: $$\frac{\alpha}{2\beta} = \frac{169}{4}$$

Coefficient of $$\sin^{-1}\left(\frac{12}{13}\right)$$: $$\frac{\alpha}{\beta} = \frac{169}{2}$$

Constant term: $$-\frac{65}{2}$$ matches.

From $$\frac{\alpha}{\beta} = \frac{169}{2}$$, and $$\frac{\alpha}{2\beta} = \frac{1}{2} \cdot \frac{\alpha}{\beta} = \frac{1}{2} \cdot \frac{169}{2} = \frac{169}{4}$$, which matches the coefficient of $$\pi$$. Thus, $$\frac{\alpha}{\beta} = \frac{169}{2}$$, so $$\alpha = 169$$, $$\beta = 2$$. These are coprime since $$169 = 13^2$$ and $$2$$ have no common factors.

Therefore, $$\alpha + \beta = 169 + 2 = 171$$.