To solve the equation \( x^2 + y^2 - 8x + 2y - 8 = 0 \), we need to rewrite it in the standard form of a circle equation, which is \( (x - h)^2 + (y - k)^2 = r^2 \). Here, \((h, k)\) is the center of the circle, and \(r\) is the radius.

### Step-by-Step Solution:

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

2. **Complete the square for the \(x\) terms:**
   - Take the coefficient of \(x\), which is \(-8\), divide by 2, and square it:
     \[
     \left(\frac{-8}{2}\right)^2 = 16
     \]
   - Add and subtract 16 inside the equation:
     \[
     x^2 - 8x + 16
     \]

3. **Complete the square for the \(y\) terms:**
   - Take the coefficient of \(y\), which is \(2\), divide by 2, and square it:
     \[
     \left(\frac{2}{2}\right)^2 = 1
     \]
   - Add and subtract 1 inside the equation:
     \[
     y^2 + 2y + 1
     \]

4. **Rewrite the equation with these squares:**
   \[
   (x^2 - 8x + 16) + (y^2 + 2y + 1) = 8 + 16 + 1
   \]

5. **Simplify the equation:**
   \[
   (x - 4)^2 + (y + 1)^2 = 25
   \]

### Conclusion:

The equation \((x - 4)^2 + (y + 1)^2 = 25\) represents a circle with:
- **Center:** \((4, -1)\)
- **Radius:** \(5\) (since \(\sqrt{25} = 5\))

Would you like more details or have any questions?

### Related Questions:

1. How do you derive the general form of the equation of a circle?
2. What are the steps to complete the square for a quadratic expression?
3. How can you find the center and radius of a circle from its equation?
4. What is the geometric interpretation of completing the square?
5. How can you verify if a given point lies on the circle defined by \((x - 4)^2 + (y + 1)^2 = 25\)?

### Tip:

When completing the square, remember to add and subtract the same value inside the equation to maintain equality. This technique is crucial for converting quadratic equations into a more manageable form.

Explore how to transform x² + y² - 8x + 2y - 8 = 0 into the standard form of a circle equation through completing the square. This solution finds the center (4, -1) and radius 5 of the circle, using key algebraic techniques suitable for high school students.

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