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Circle ABC lies on a cone

Short answer

The circle ABC is intersection of plane x/a+y/b+z/c=λ and sphere x^2+y^2+z^2-λ(ax+by+cz)=0. Eliminating λ yields the cone (b/c+c/b)yz+(c/a+a/c)zx+(a/b+b/a)xy=0.

A variable plane is parallel to the plane x/a + y/b + z/c = 1 and meets the axes in A, B, C. Prove that the circle ABC lies on the cone (b/c + c/b)yz + (c/a + a/c)zx + (a/b + b/a)xy = 0.

The circle ABC is the intersection of the variable plane with the sphere through O, A, B, C. Eliminating the variable parameter between the two gives a cone with vertex at the origin.

Plane and its intercepts

A plane parallel to x/a + y/b + z/c = 1 is x/a + y/b + z/c = λ, where λ is the variable. It meets the axes at

A(aλ, 0, 0), B(0, bλ, 0), C(0, 0, cλ).

Sphere through O, A, B, C

A sphere through the origin has the form x² + y² + z² + 2ux + 2vy + 2wz = 0. Putting A, B, C gives 2u = −aλ, 2v = −bλ, 2w = −cλ, so

x² + y² + z² − λ(ax + by + cz) = 0          ... (1)

The circle ABC

The circle is given by (1) together with

x/a + y/b + z/c = λ                          ... (2)

Eliminate λ

Substitute λ from (2) into (1) to make it homogeneous:

x² + y² + z² − (ax + by + cz)(x/a + y/b + z/c) = 0.

Expand the product:

(ax+by+cz)(x/a+y/b+z/c) = x² + y² + z²
   + xy(a/b + b/a) + yz(b/c + c/b) + zx(c/a + a/c)

The squares cancel, leaving

(b/c + c/b)yz + (c/a + a/c)zx + (a/b + b/a)xy = 0.

Every point of the circle satisfies this for every λ, so the circle ABC lies on this cone (a homogeneous equation of second degree, vertex at O).

✓ Answer checked by Claude Opus 5.5 (high effort)

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