To find the center of a circle given two points on its circumference and the radius, follow these steps:

1. Find the midpoint of the segment connecting the given points: The midpoint formula is: $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ For the points $(-2, 0)$ and $(8, 0)$: $\left( \frac{-2 + 8}{2}, \frac{0 + 0}{2} \right) = (3, 0)$ So, the midpoint is $(3, 0)$.

2. Determine the distance from the midpoint to one of the points: This is half the length of the segment connecting the given points. The distance formula is: $\text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ For the points $(-2, 0)$ and $(8, 0)$: $\text{distance} = \sqrt{(8 - (-2))^2 + (0 - 0)^2} = \sqrt{10^2} = 10$ So, half the distance is $5$.

3. Find the points that are at the given radius from the midpoint: Since the distance from the midpoint to each center must be perpendicular to the segment joining the two points and have the length of the radius. The coordinates of these points are: $(3 \pm \sqrt{5^2 - 5^2}, 0 \pm 5) = (3, 5) \text{ or } (3, -5)$ The centers are $(3, 5)$ and $(3, -5)$.

Therefore, the possible centers of the circle are $(3, 5)$ and $(3, -5)$.

Would you like more details or have any questions about this solution?

Here are some related questions for further practice:

1. How would you find the center if the radius was different, say 7?
2. How do you derive the equation of the circle given the center and radius?
3. What if the points were not on the x-axis, how would you approach the problem?
4. Can you find the equation of the line passing through the points $(-2,0)$ and $(8,0)$?
5. How would you prove geometrically that there are only two possible centers for the circle?
6. What are the general properties of a circle and how do they apply here?
7. How does the distance formula help in solving problems related to circles?
8. What is the role of the perpendicular bisector in finding the center of a circle?

Tip: When dealing with geometric problems, sketching a diagram can often provide valuable insights and simplify the problem-solving process.