--- title: "Sphere through circle touching a plane" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:sphere-through-circle-touching-plane" source: https://www.bctengineer.com/q/semester_2/mathematics/sphere-through-circle-touching-plane site: BCT Engineering --- # Sphere through circle touching a plane Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2073 Magh, 5 marks - Video: [ENGINEERING MATHS 2 - UNIT 5 (SPHERE) - TYPE 3 - Sphere passing through the circle](https://www.youtube.com/watch?v=92CB1MEjm6Q) — VED ACADEMY - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/sphere-through-circle-touching-plane - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question Find the equation of spheres passing through the circle x²+y²+z²−6x−2z+5 = 0, y = 0 and touching the plane 3y+4z+5 = 0. ## Short answer Family of spheres: x²+y²+z²-6x-2z+5+λy=0. Centre (3, -λ/2, 1), radius²=5+λ²/4. Tangency condition |3(-λ/2)+4(1)+5|/5 = radius gives λ=-4 and λ=-11/4. Spheres: x²+y²+z²-6x-4y-2z+5=0 and 4x²+4y²+4z²-24x-11y-8z+20=0. ## Answer **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: ```text 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 ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the family of spheres through the circle? A. x^2 + y^2 + z^2 - 6x - 2z + 5 + lambda y = 0 B. x^2 + y^2 + z^2 - 6x - 2z + 5 + lambda = 0 C. x^2 + y^2 + z^2 - 6x - 2z + 5 = lambda y^2 D. x^2 + y^2 + z^2 - 6x - 2z + 5 + lambda x = 0 Answer: A. The circle lies in y = 0, so adding lambda y keeps it on every sphere. (Key point: Family of spheres) 2. What are the centre and the radius squared? A. Centre (3, lambda/2, 1), radius^2 = 5 + lambda^2/4 B. Centre (3, -lambda/2, 1), radius^2 = 5 - lambda^2/4 C. Centre (3, -lambda/2, 1), radius^2 = 5 + lambda^2/4 D. Centre (-3, lambda/2, -1), radius^2 = 5 + lambda Answer: C. r^2 = 9 + lambda^2/4 + 1 - 5. (Key point: Centre and radius) 3. What condition gives the quadratic in lambda? A. The plane passes through the centre B. Distance from the centre to the plane equals the radius C. The centre lies on the plane y = 0 D. The radius equals 1 Answer: B. d^2 = r^2 gives 4 lambda^2 + 27 lambda + 44 = 0. (Key point: Tangency) 4. Which values of lambda are found? A. lambda = -4 and lambda = -11/4 B. lambda = 4 and lambda = 11/4 C. lambda = -4 only D. lambda = -11 and lambda = -4 Answer: A. (lambda + 4)(4 lambda + 11) = 0, giving two spheres. 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