Right circular cylinder with guiding circle x²+y²+z²=9

A right circular cylinder with a given guiding circle has its axis perpendicular to the plane of the circle and passing through the centre of the circle. Its radius is the radius of the circle.

Circle: x² + y² + z² = 9 (sphere, centre O, radius 3), plane x − y + z = 3.

Centre and radius of the circle

The centre is the foot of the perpendicular from O on the plane, along the normal (1, −1, 1): the point (t, −t, t) with t + t + t = 3, so t = 1. Centre C = (1, −1, 1).

Distance from O to the plane = 3/√3 = √3, so

r² = 9 − 3 = 6.

Axis

The axis passes through C (and O) with direction ratios (1, −1, 1), the normal of the plane.

Equation of the cylinder

Using the distance of P(x, y, z) from the axis through O with direction (1, −1, 1):

(distance)² = OP² − (projection on axis)²
 x² + y² + z² − (x − y + z)²/3 = 6

Multiply by 3:

3(x²+y²+z²) − (x² + y² + z² − 2xy + 2xz − 2yz) = 18
2x² + 2y² + 2z² + 2xy + 2yz − 2zx = 18

Answer: x² + y² + z² + xy + yz − zx = 9.

Check: the point (1+√3, −1+√3, 1) is on the guiding circle (it is on the plane x − y + z = 3 and its squares add to 9). It satisfies x² + y² + z² + xy + yz − zx = 9.