Tangent plane to sphere through line

Sphere. x² + y² + z² + 6x − 2z + 1 = 0 has centre C(−3, 0, 1) and radius r = √(9 + 0 + 1 − 1) = 3.

Line. 3(16 − x) = 3z = 2y + 30 is the intersection of the planes x + z − 16 = 0 (from 3(16 − x) = 3z) and 3z − 2y − 30 = 0.

Planes through the line

(x + z − 16) + k(−2y + 3z − 30) = 0
x − 2k y + (1 + 3k) z − (16 + 30k) = 0

Tangent condition

The distance from C to the plane must equal the radius 3. For the plane with coefficients A = 1, B = −2k, C₁ = 1 + 3k and D = −(16 + 30k):

A(−3) + B(0) + C₁(1) + D = −3 + 1 + 3k − 16 − 30k = −18 − 27k

Condition: (−18 − 27k)² = 9 [1 + 4k² + (1 + 3k)²]

324 + 972k + 729k² = 9 (2 + 6k + 13k²) = 18 + 54k + 117k²
612k² + 918k + 306 = 0
2k² + 3k + 1 = 0
(2k + 1)(k + 1) = 0  →  k = −1/2 or k = −1

The planes

• k = −1: x + 2y − 2z + 14 = 0, i.e. x + 2y − 2z + 14 = 0
• k = −1/2: x + y − z/2 − 1 = 0, i.e. 2x + 2y − z − 2 = 0

Check: the point (16, −15, 0) lies on the line and on both planes (16 − 30 + 14 = 0; 32 − 30 − 2 = 0), and the distance from C to each plane is 9/3 = 3 = radius.

Answer: two tangent planes, x + 2y − 2z + 14 = 0 and 2x + 2y − z − 2 = 0.