Find the equation of the circle that has the center (2,0) and passes through (-1,-5)

Solution:
1. Given:
- Center of the circle: (2, 0)
- Point on the circle: (-1, -5)

2. Use the standard equation of a circle:
- (x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center and r is the radius.

3. Substitute the center into the equation:
- (x - 2)^2 + (y - 0)^2 = r^2
- Simplifies to: (x - 2)^2 + y^2 = r^2

4. Substitute point (-1, -5) to find r^2:
- ((-1) - 2)^2 + ((-5) - 0)^2 = r^2
- (-3)^2 + (-5)^2 = r^2
- 9 + 25 = r^2
- r^2 = 34

5. Therefore, the equation of the circle is:
- (x - 2)^2 + y^2 = 34