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Determine the area inside the circle defined by: x2 +y2 −8x+4y+4=0

but outside the triangle bounded by the three lines below:

y=x−7

y = 51(2x−29) x=7

Let's break down Problem 9 into its parts for a solution:

### Step 1: **Equation of the Circle**

The equation given is:

\[
x^2 + y^2 - 8x + 4y + 4 = 0
\]

We can rewrite this equation in standard form by completing the square.

1. **Group the \(x\) and \(y\) terms**:
   \[
   (x^2 - 8x) + (y^2 + 4y) = -4
   \]

2. **Complete the square**:
   - For \(x^2 - 8x\), take half of \(-8\), square it, and add 16:
     \[
     (x^2 - 8x + 16)
     \]
   - For \(y^2 + 4y\), take half of 4, square it, and add 4:
     \[
     (y^2 + 4y + 4)
     \]

   Now add the same constants to the other side of the equation:
   \[
   (x - 4)^2 + (y + 2)^2 = 16
   \]

This is the equation of a circle with:
- Center at \((4, -2)\)
- Radius \(4\) (since \(\sqrt{16} = 4\))

### Step 2: **Triangle Bounded by the Lines**

The three lines are:
1. \(y = x - 7\)
2. \(y = \frac{1}{5}(2x - 29)\)
3. \(x = 7\)

These lines form a triangle in the coordinate plane.

- **\(x = 7\)** is a vertical line passing through \(x = 7\).
- **\(y = x - 7\)** is a line with slope 1 and y-intercept \(-7\).
- **\(y = \frac{1}{5}(2x - 29)\)** is a line with slope \(\frac{2}{5}\) and y-intercept \(\frac{-29}{5}\).

We need to find the points where these lines intersect to form the vertices of the triangle.

### Step 3: **Determine the Area**

We need to calculate the area inside the circle but outside this triangle. This requires:
- Finding the area of the region of the circle that overlaps with the triangle.
- Subtracting the area of the triangle from the area of the circle.

I'll compute these areas step by step.The three points of intersection between the lines form the vertices of the triangle:

1. Intersection of \(y = x - 7\) and \(y = \frac{1}{5}(2x - 29)\) gives \((2, -5)\).
2. Intersection of \(y = x - 7\) and \(x = 7\) gives \((7, 0)\).
3. Intersection of \(y = \frac{1}{5}(2x - 29)\) and \(x = 7\) gives \((7, -3)\).

### Step 4: **Calculate the Area of the Triangle**

Now that we have the vertices of the triangle \((2, -5)\), \((7, 0)\), and \((7, -3)\), we can calculate the area of the triangle using the formula for the area of a triangle with coordinates \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

I'll compute this now.The area of the triangle formed by the lines is \(7.5\) square units.

### Step 5: **Area of the Circle**

The area of the circle with radius \(4\) is given by the formula:

\[
\text{Area of the circle} = \pi r^2 = \pi (4)^2 = 16\pi \approx 50.27 \, \text{square units}
\]

### Step 6: **Final Area**

The problem asks for the area inside the circle but outside the triangle. To find this, we subtract the area of the triangle from the area of the circle:

\[
\text{Desired area} = 16\pi - 7.5 \approx 50.27 - 7.5 = 42.77 \, \text{square units}
\]

Thus, the area inside the circle but outside the triangle is approximately **42.77 square units**.

---

Would you like more details on any part of the solution?

Here are some related questions:
1. How is the equation of a circle derived from general form?
2. What is the process for completing the square in algebra?
3. How do you find the intersection points of two lines?
4. What are other methods to calculate the area of a triangle?
5. How do we calculate areas under curves?

**Tip**: Always sketch out geometric shapes when solving area problems to visualize the regions in question!

Determine the area inside the circle defined by: x^2 + y^2 − 8x + 4y + 4 = 0 but outside the triangle bounded by the three lines y = x - 7, y = (1/5)(2x - 29), and x = 7.

Learn how to determine the area inside the circle x^2 + y^2 − 8x + 4y + 4 = 0 but outside the triangle bounded by the lines y = x - 7, y = (1/5)(2x - 29), and x = 7. This involves concepts of circle equations, triangle geometry, and area calculations.

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