Sphere through circle touching a plane

Given. Circle: x² + y² + z² − 6x − 2z + 5 = 0, y = 0. Touching plane: 3y + 4z + 5 = 0.

Family of spheres through the circle

x² + y² + z² − 6x − 2z + 5 + λy = 0

Centre: (3, −λ/2, 1).

Radius²: r² = 9 + λ²/4 + 1 − 5 = 5 + λ²/4.

Tangency condition

Distance from the centre to the plane 3y + 4z + 5 = 0 must equal the radius:

d = |3(−λ/2) + 4(1) + 5| / 5 = |9 − 3λ/2| / 5

d² = r²
(9 − 3λ/2)² / 25 = 5 + λ²/4
81 − 27λ + 9λ²/4 = 125 + 25λ²/4

Multiply by 4: 324 − 108λ + 9λ² = 500 + 25λ², so

16λ² + 108λ + 176 = 0
4λ² + 27λ + 44 = 0
(λ + 4)(4λ + 11) = 0   →   λ = −4  or  λ = −11/4

The spheres

• λ = −4: x² + y² + z² − 6x − 4y − 2z + 5 = 0 (centre (3, 2, 1), radius 3; distance to plane = 15/5 = 3 ✓).
• λ = −11/4: 4x² + 4y² + 4z² − 24x − 11y − 8z + 20 = 0 (centre (3, 11/8, 1), radius 21/8; distance to plane = (105/8)/5 = 21/8 ✓).

Answer: two spheres, as above.