A sphere of radius 3 centered at (1,-1,a) where a>0, contains the point (0,0,1). Write the equation of the sphere

Solution:

1. Given:
- Radius: 3
- Center: (1, -1, a)
- Contains the point: (0, 0, 1)

2. Use the distance formula to find the relationship between a and the given point:
\sqrt{(0 - 1)^2 + (0 + 1)^2 + (1 - a)^2} = 3

3. Simplify the distance equation:
\sqrt{1 + 1 + (1 - a)^2} = 3

4. Square both sides to remove the square root:
1 + 1 + (1 - a)^2 = 9

5. Simplify the equation:
2 + (1 - a)^2 = 9
(1 - a)^2 = 7
1 - a = \pm \sqrt{7}

6. Solve for a:
- Case 1: 1 - a = \sqrt{7}
a = 1 - \sqrt{7}

- Case 2: 1 - a = -\sqrt{7}
a = 1 + \sqrt{7}

Since a > 0, we choose a = 1 + \sqrt{7} .

7. Write the equation of the sphere:
- The general formula for the equation of a sphere with center (h, k, l) and radius $r is: (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 - Substitute h = 1 , k = -1 , l = a = 1 + \\sqrt{7} , and r = 3 : (x - 1)^2 + (y + 1)^2 + (z - (1 + \\sqrt{7}))^2 = 9 Thus, the equation of the sphere is: (x - 1)^2 + (y + 1)^2 + (z - (1 + \\sqrt{7}))^2 = 9 $$