--- title: "Circle ABC lies on a cone" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:circle-abc-lies-cone" source: https://www.bctengineer.com/q/semester_2/mathematics/circle-abc-lies-cone site: BCT Engineering --- # Circle ABC lies on a cone Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2069 Poush, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/circle-abc-lies-cone - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question 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. ## 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. ## Answer 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 ```text x² + y² + z² − λ(ax + by + cz) = 0 ... (1) ``` #### The circle ABC The circle is given by (1) together with ```text 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: ```text (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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A plane x/a + y/b + z/c = λ meets the axes at: A. A(aλ, 0, 0), B(0, bλ, 0), C(0, 0, cλ) B. A(a/λ, 0, 0), ... C. A(a, 0, 0), B(0, b, 0), C(0, 0, c) D. A(λ/a, 0, 0), ... Answer: A. Put y = z = 0 etc. to get x = aλ. (Key point: Intercepts) 2. The sphere through the origin and A, B, C is: A. x^2+y^2+z^2 + λ(ax+by+cz) = 0 B. x^2+y^2+z^2 - λ(ax+by+cz) = 0 C. x^2+y^2+z^2 - λ(x/a+y/b+z/c) = 0 D. x^2+y^2+z^2 - λ^2 = 0 Answer: B. 2u = -aλ etc. (Key point: Sphere through O, A, B, C) 3. How is a cone obtained? A. Set λ = 0 B. Add the plane and sphere equations C. Replace λ in the sphere by x/a + y/b + z/c from the plane, making a homogeneous equation D. Differentiate with respect to λ Answer: C. A homogeneous second-degree equation represents a cone with vertex at O. (Key point: Eliminating the parameter) 4. In the expansion of (ax+by+cz)(x/a+y/b+z/c), what happens to the square terms? A. They give x^2+y^2+z^2, which cancels the sphere's squares B. They give 2(x^2+y^2+z^2) C. They vanish by themselves D. They give abc(x^2+y^2+z^2) Answer: A. Only cross terms like (a/b + b/a)xy remain. 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