--- title: "Sphere with circle as great circle" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:sphere-circle-as-great-circle" source: https://www.bctengineer.com/q/semester_2/mathematics/sphere-circle-as-great-circle site: BCT Engineering --- # Sphere with circle as great circle Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2068 Bhadra, 5 marks - Video: [Sphere[Great circle]|Lecture-2|Mathmatics-II BE all questions solution TU PU KU PoU #mathematics](https://www.youtube.com/watch?v=Ir7fAnN7Tpg) — Civil Engineering Solution Channel - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/sphere-circle-as-great-circle - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question Find the equation of the sphere through the circle x²+y²+z²−3x+4y−2z−5 = 0, 5x−2y+4z+7 = 0 as a great circle. ## Short answer The sphere is from S + kP = 0 with centre ((3-5k)/2,(2k-4)/2,(2-4k)/2). Setting the centre on the plane gives k=1, resulting in x^2+y^2+z^2+2x+2y+2z+2=0. ## Answer A circle is a great circle of a sphere when the sphere's centre lies on the plane of the circle. The circle is S = x² + y² + z² − 3x + 4y − 2z − 5 = 0, P = 5x − 2y + 4z + 7 = 0. #### Family of spheres S + kP = 0: x² + y² + z² + (5k − 3)x + (4 − 2k)y + (4k − 2)z + 7k − 5 = 0. Centre = ( (3 − 5k)/2, (2k − 4)/2, (2 − 4k)/2 ). #### Centre on the plane P = 0 ```text 5(3−5k)/2 − 2(2k−4)/2 + 4(2−4k)/2 + 7 = 0 (15 − 25k) − (4k − 8) + (8 − 16k) + 14 = 0 45 − 45k = 0 → k = 1 ``` (Check: 15 − 25k + 8 − 4k + 8 − 16k + 14 = 45 − 45k.) #### Required sphere Put k = 1: x² + y² + z² + 2x + 2y + 2z + 2 = 0 Its centre is (−1, −1, −1) and 5(−1) − 2(−1) + 4(−1) + 7 = 0, so the centre is on the plane. **Answer: x² + y² + z² + 2x + 2y + 2z + 2 = 0** (centre (−1, −1, −1), radius 1). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. To make the given circle a great circle: A. Choose k so the centre of S + kP lies on P = 0 B. Choose k so S + kP passes through the origin C. Choose k = 0 D. Choose k so the radius is 1 Answer: A. The centre being in the plane of the circle makes the circle a great circle. (Key point: Condition) 2. Centre of S + kP = 0 with the x coefficient 5k - 3 is: A. (5k - 3)/2 B. x = (3 - 5k)/2 and so on C. 3 - 5k D. (5k - 3) Answer: B. Centre = (-1/2)(coefficients). (Key point: Centre) 3. Substituting the centre into 5x - 2y + 4z + 7 = 0 gives: A. k = 0 B. k = 45 C. k = 1 D. k = -1 Answer: C. 45 - 45k = 0. (Key point: Value of k) 4. The sphere is: A. x^2+y^2+z^2 + 2x + 2y + 2z + 2 = 0 B. x^2+y^2+z^2 - 2x - 2y - 2z + 2 = 0 C. x^2+y^2+z^2 + 2x + 2y + 2z - 2 = 0 D. x^2+y^2+z^2 + 8x + 2y + 6z + 2 = 0 Answer: A. k = 1: x coefficient 2, y coefficient 2, z coefficient 2, constant 2. 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