The area (in sq. units) of the part of circle x² + y² = 169 which is below the line 5x - y = 13 is (

Q: Wait, why do we integrate with respect to $y$ instead of $x$?

Great question! If you integrate with respect to $x$, the upper boundary changes at $x=5$ (from the circle to the line), forcing you to split the integral into two distinct parts. By integrating with respect to $y$, the boundary is a single, continuous function of $y$ across the entire interval, which simplifies the calculation significantly.

Q: How do we handle the lower limit of $-13$ in the $\sin^{-1}$ term?

When you substitute $y = -13$ into the term $\sin^{-1}(\frac{y}{13})$, you get $\sin^{-1}(-1)$. Recall that $\sin^{-1}(-1) = -\frac{\pi}{2}$. This is a standard trigonometric value that appears frequently in these types of problems, so keep it in your toolkit!

Q: Why is the line equation $x = \frac{y+13}{5}$?

We are integrating with respect to $y$, so we need to express $x$ as a function of $y$. Starting from the given line equation $5x - y = 13$, we simply add $y$ to both sides to get $5x = y + 13$, and then divide by $5$ to isolate $x$.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with a circle defined by the equation $x^2 + y^2 = 169$, which is a symmetric entity with a radius of $r=13$ centered at the origin. We also have a line defined by $5x - y = 13$ that intersects this circle.

Our goal is to find the area of the region trapped below this line but within the circle. To achieve this, we must first determine the points of intersection between these two geometric figures.

### The Strategic Choice of Integration

While integrating with respect to $x$ would require splitting the region at $x=5$, integrating with respect to $y$ allows us to define the area as a single, continuous strip. We define the area as the integral of the horizontal distance between the circle and the line:

$$Area = \int_{y_1}^{y_2} (x_{	ext{circle}} - x_{	ext{line}}) dy$$

By substituting $x = \frac{y+13}{5}$ into the circle equation $x^2 + y^2 = 169$, we solve for the intersection points. This yields the limits of integration as $y_1 = -13$ and $y_2 = 12$.

### The Heavy Lifting: The Circular Integral

We now evaluate the integral of the circular arc, $\int_{-13}^{12} \sqrt{169 - y^2} dy$. Using the standard integral formula:

$$\int \sqrt{a^2 - y^2} dy = \frac{y}{2}\sqrt{a^2 - y^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}ight)$$

Setting $a=13$ and evaluating from $-13$ to $12$, we find the value at the upper limit $y=12$:

$$\left[ \frac{12}{2}\sqrt{169 - 144} + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}ight) ight] = 30 + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}ight)$$

At the lower limit $y=-13$, the square root term vanishes, leaving $\frac{169}{2}\sin^{-1}(-1) = -\frac{169\pi}{4}$. Subtracting these values gives the area under the circular arc.

### The Linear Counterpart and Final Synthesis

Next, we evaluate the linear component:

$$\int_{-13}^{12} \frac{y+13}{5} dy = \frac{1}{5} \left[ \frac{y^2}{2} + 13y ight]_{-13}^{12} = 62.5$$

Combining the circular area and the linear area, we arrive at the expression:

$$Area = \frac{169\pi}{4} - 32.5 + \frac{169}{2}\sin^{-1}\left(\frac{12}{13}ight)$$

Comparing this to the target form $\frac{\pi \alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta} \sin^{-1}(\frac{12}{13})$, we identify $\alpha = 169$ and $\beta = 2$. Since these are coprime, the final result is:

**$$\alpha + \beta = 171$$**

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Finding Intersection Points

Solving the Quadratic Equation

Defining the Integration Strategy

Setting up the Integral

Evaluating the Linear Integral

The Circular Part: Integration Formula

Applying the Upper Limit

Applying the Lower Limit

Combining the Results

Final Algebraic Form

Identifying $α$ and $β$

The Final Sum

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Strategic Choice of Integration

The Heavy Lifting

The Linear Counterpart and Final Synthesis

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