Given circle and line

The circle is $x^2+y^2=169$ so its center is $(0,0)$ and radius is $r=13.$

The line is $5x-y=13 \implies y=5x-13.$

We need the area of the part of the circle lying below this line.

Interpret the required region

The line cuts the circle into two segments. Since the origin $(0,0)$ satisfies $5(0)-0=0<13,$ the center lies on the side $5x-y<13 \iff y>5x-13,$ so the center is above the line.

Therefore, the part of the circle below the line is the minor segment cut off by the chord.

Distance of the line from the center

Write the line as $5x-y-13=0.$ Distance from $(0,0)$ to this line is $d=\frac{| -13 |}{\sqrt{5^2+(-1)^2}}=\frac{13}{\sqrt{26}}.$

Area of the minor segment

If a chord is at distance $d$ from the center of a circle of radius $r$, then the area of the minor segment is $A=r^2\cos^{-1}\left(\frac{d}{r}\right)-d\sqrt{r^2-d^2}.$

Here, $\frac{d}{r}=\frac{13/\sqrt{26}}{13}=\frac{1}{\sqrt{26}}.$ Hence $A=169\cos^{-1}\left(\frac{1}{\sqrt{26}}\right)-\frac{13}{\sqrt{26}}\sqrt{169-\frac{169}{26}}.$

Simplify the second term

$169-\frac{169}{26}=169\left(1-\frac{1}{26}\right)=169\cdot\frac{25}{26}.$ So $\sqrt{169-\frac{169}{26}}=13\cdot\frac{5}{\sqrt{26}}=\frac{65}{\sqrt{26}}.$

Therefore, $d\sqrt{r^2-d^2}=\frac{13}{\sqrt{26}}\cdot\frac{65}{\sqrt{26}}=\frac{845}{26}=\frac{65}{2}.$

So, $A=169\cos^{-1}\left(\frac{1}{\sqrt{26}}\right)-\frac{65}{2}.$

Convert the inverse cosine

Let $\theta=\cos^{-1}\left(\frac{1}{\sqrt{26}}\right)$. Then $\cos\theta=\frac{1}{\sqrt{26}}, \qquad \sin\theta=\frac{5}{\sqrt{26}}.$

Since $\tan\theta=5,$ we may write $\theta=\frac{\pi}{2}-\sin^{-1}\left(\frac{1}{\sqrt{26}}\right).$

But the expression in the question involves $\sin^{-1}(12/13)$, so let us use the intersection geometry directly.

Find points of intersection to identify the angle

Solve $x^2+(5x-13)^2=169.$ This gives $x^2+25x^2-130x+169=169$ $26x^2-130x=0$ $26x(x-5)=0.$

So intersection points are $(0,-13), \quad (5,12).$

These correspond to radii from the origin to those points. Their polar angles are:

• for $(0,-13)$: angle $=-\frac{\pi}{2}$,
• for $(5,12)$: since $\sin\phi=\frac{12}{13}, \quad \cos\phi=\frac{5}{13},$ we have $\phi=\sin^{-1}\left(\frac{12}{13}\right).$

Thus the central angle subtending the minor segment is $\theta=\frac{\pi}{2}+\sin^{-1}\left(\frac{12}{13}\right).$

Hence the segment area is also $A=\frac{r^2}{2}(\theta-\sin\theta).$ But more directly, since we already found the triangular part contributes $\frac{65}{2}$, we get $A=\frac{169}{2}\left(\frac{\pi}{2}+\sin^{-1}\left(\frac{12}{13}\right)\right)-\frac{65}{2}.$

Therefore, $A=\frac{169\pi}{4}-\frac{65}{2}+\frac{169}{2}\sin^{-1}\left(\frac{12}{13}\right).$

Match with the given form

Given $A=\frac{\pi\alpha}{2\beta}-\frac{65}{2}+\frac{\alpha}{\beta}\sin^{-1}\left(\frac{12}{13}\right).$

Comparing, $\frac{\pi\alpha}{2\beta}=\frac{169\pi}{4}, \qquad \frac{\alpha}{\beta}=\frac{169}{2}.$

Since $169$ and $2$ are coprime, $\alpha=169, \quad \beta=2.$

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

Comparison with stored answer

Derived answer = $171$.

Stored correct answer = $171$.

They agree.