Sphere through circle meeting plane in circle of radius 3

Given. Circle: x² + y² = 4, z = 0. The sphere also meets the plane x + 2y + 2z = 0 in a circle of radius 3.

Family of spheres through the circle

The circle lies on the sphere x² + y² + z² = 4 and on the plane z = 0, so every sphere through it is

x² + y² + z² − 4 + kz = 0.

Centre: (0, 0, −k/2). Radius²: R² = 4 + k²/4.

Section by the plane x + 2y + 2z = 0

Distance of the centre from the plane:

p = |0 + 0 + 2(−k/2)| / √(1 + 4 + 4) = |k| / 3

The circle of section has radius 3, so

R² − p² = 9
4 + k²/4 − k²/9 = 9
(5/36) k² = 5
k² = 36   →   k = 6 or k = −6

Spheres

• k = 6: x² + y² + z² + 6z − 4 = 0
• k = −6: x² + y² + z² − 6z − 4 = 0

Check for k = 6: centre (0, 0, −3), R² = 13, p = 2, R² − p² = 9 ✓.

Answer: x² + y² + z² ± 6z − 4 = 0.