Centre and radius of a circle on sphere

The circle is the intersection of the sphere S: x² + y² + z² + x + y + z − 4 = 0 and the plane x + y + z = 0.

Sphere

Compare with x² + y² + z² + 2ux + 2vy + 2wz + d = 0: u = v = w = 1/2, d = −4.

Centre C = (−1/2, −1/2, −1/2)
Radius R = √(u² + v² + w² − d) = √(3/4 + 4) = √19 / 2

Distance from C to the plane

Plane x + y + z = 0, normal (1, 1, 1):

p = |−1/2 − 1/2 − 1/2| / √3 = (3/2)/√3 = √3/2

Radius of the circle

r = √(R² − p²) = √(19/4 − 3/4) = √4 = 2

Centre of the circle

It is the foot of the perpendicular from C to the plane. Points on the perpendicular: (−1/2 + t, −1/2 + t, −1/2 + t). Put in x + y + z = 0:

−3/2 + 3t = 0, t = 1/2. The point is (0, 0, 0).

Answer: centre (0, 0, 0), radius 2.