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Sphere through circle of two spheres, centre on a plane

Short answer

Using the sphere pencil S1 + kS2 = 0 and imposing the centre lies on 4x-5y-z-3=0 gives k = -3. Simplifying yields x²+y²+z²+7x+9y-11z-1 = 0.

Show that the equation of the sphere through the circle x²+y²+z²−2x−3y+4z+8 = 0 and x²+y²+z²+4x+5y−6z+2 = 0 and having its centre on the plane 4x−5y−z−3 = 0 is x²+y²+z²+7x+9y−11z−1 = 0. [First two equations are the two spheres whose intersection is the circle, as read from scan]

[The two given equations are the two spheres whose common circle is meant (as read from the scan). The given answer is verified below.]

Let S₁ = x² + y² + z² − 2x − 3y + 4z + 8 and S₂ = x² + y² + z² + 4x + 5y − 6z + 2.

Family of spheres through the circle

S₁ + kS₂ = 0:

(1+k)(x²+y²+z²) + (4k−2)x + (5k−3)y + (4−6k)z + (8+2k) = 0

Centre = ( (1−2k)/(1+k), (3−5k)/(2(1+k)), (3k−2)/(1+k) ).

Centre on the plane 4x − 5y − z − 3 = 0

Multiply by 2(1 + k):

8(1−2k) − 5(3−5k) − 2(3k−2) − 6(1+k) = 0
8 − 16k − 15 + 25k − 6k + 4 − 6 − 6k = 0
−9 − 3k = 0   →   k = −3

The sphere

S₁ − 3S₂ = 0:

−2(x²+y²+z²) − 14x − 18y + 22z + 2 = 0

Divide by −2:

x² + y² + z² + 7x + 9y − 11z − 1 = 0, which is the required result.

Check: the centre is (−7/2, −9/2, 11/2); 4(−7/2) − 5(−9/2) − 11/2 − 3 = −14 + 22.5 − 5.5 − 3 = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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