Plane touches sphere, find point of contact

Sphere. x² + y² + z² − 4x + 2y − 4 = 0. Comparing with x² + y² + z² + 2ux + 2vy + 2wz + d = 0: u = −2, v = 1, w = 0, d = −4.

Centre C = (2, −1, 0)
Radius r = √(4 + 1 + 0 + 4) = 3

Proof of tangency

A plane touches a sphere when the perpendicular distance from the centre to the plane equals the radius. Plane: 2x − y + 2z − 14 = 0, with |normal| = √(4 + 1 + 4) = 3.

d = |2(2) − (−1) + 2(0) − 14| / 3 = |4 + 1 − 14| / 3 = 9/3 = 3

This equals the radius 3, so the plane touches the sphere. ∎

Point of contact

It is the foot of the perpendicular from C to the plane. The perpendicular from C has direction (2, −1, 2):

(2 + 2t, −1 − t, 2t).

Put on the plane:

2(2 + 2t) − (−1 − t) + 2(2t) − 14 = 0
4 + 4t + 1 + t + 4t − 14 = 0
9t = 9   →  t = 1

Point of contact: (4, −2, 2).

Check: on the sphere: 16 + 4 + 4 − 16 − 4 − 4 = 0 ✓. On the plane: 8 + 2 + 4 − 14 = 0 ✓.

Answer: the point of contact is (4, −2, 2).