Right circular cylinder with guiding circle

Given. Guiding circle: sphere S: x² + y² + z² − x − y − z = 0 and plane x + y + z = 1.

A right circular cylinder has its axis perpendicular to the plane of its guiding circle, passing through the centre of the circle, and its radius equals the circle's radius.

Axis

The plane's normal is (1, 1, 1), so the axis has direction ratios (1, 1, 1).

Sphere: centre (1/2, 1/2, 1/2), radius R = √(3/4) = √3/2.

Distance from the sphere's centre to the plane: p = |3/2 − 1|/√3 = 1/(2√3), so p² = 1/12.

Circle radius: r² = R² − p² = 3/4 − 1/12 = 2/3.

The circle's centre is the foot of the perpendicular from (1/2, 1/2, 1/2): points (1/2 + t, 1/2 + t, 1/2 + t) with 3/2 + 3t = 1, t = −1/6. Centre = (1/3, 1/3, 1/3).

So the axis is (x − 1/3)/1 = (y − 1/3)/1 = (z − 1/3)/1.

Equation of the cylinder

A point P(x, y, z) lies on the cylinder if its distance from the axis is r:

|PA|² − (PA·n)²/|n|² = r², with A = (1/3, 1/3, 1/3), n = (1, 1, 1).

Σ(x − 1/3)² − (x + y + z − 1)²/3 = 2/3
x² + y² + z² − (2/3)s + 1/3 − (s² − 2s + 1)/3 = 2/3      (s = x + y + z)
3(x² + y² + z²) − 2s + 1 − s² + 2s − 1 = 2
3(x² + y² + z²) − (x + y + z)² = 2
2(x² + y² + z²) − 2(xy + yz + zx) = 2

Answer: x² + y² + z² − xy − yz − zx = 1.