Sphere with given circle as great circle

A sphere has a given circle as a great circle when the circle lies on the sphere and the centre of the sphere lies in the plane of the circle.

The circle is x² + y² + z² = 9, x − 2y + 2z = 5.

Equation of the family

Every sphere through the circle is

S + kP = 0, i.e. x² + y² + z² − 9 + k(x − 2y + 2z − 5) = 0.

Rewrite it as

x² + y² + z² + kx − 2ky + 2kz − 9 − 5k = 0.

Its centre is (−k/2, k, −k).

Centre on the plane

The circle is a great circle, so the centre must lie on the plane x − 2y + 2z = 5:

(−k/2) − 2(k) + 2(−k) = 5
−9k/2 = 5   →   k = −10/9

Required sphere

x² + y² + z² − 9 − (10/9)(x − 2y + 2z − 5) = 0
9(x² + y² + z²) − 81 − 10x + 20y − 20z + 50 = 0

Answer: 9(x² + y² + z²) − 10x + 20y − 20z − 31 = 0.

Check: the centre is (5/9, −10/9, 10/9), and 5/9 + 20/9 + 20/9 = 5, so it lies on the plane.