--- title: "Tangent plane to sphere through line" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:tangent-plane-sphere-through-line" source: https://www.bctengineer.com/q/semester_2/mathematics/tangent-plane-sphere-through-line site: BCT Engineering --- # Tangent plane to sphere through line Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2075 Baisakh, 5 marks; 2072 Ashwin, 5 marks; 2070 Bhadra, 5 marks (asked 3 times) - Video: [Equation of tangent plane to the sphere|| tangent plane to sphere||solid geometry.](https://www.youtube.com/watch?v=_WNva4-q7bA) — Maths class 369 - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/tangent-plane-sphere-through-line - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question Obtain the equation of tangent plane to the sphere x²+y²+z²+6x−2z+1 = 0 which passes through the line 3(16−x) = 3z = 2y+30. ## Short answer Plane through line: x−2ky+(1+3k)z−(16+30k)=0. Distance from center (-3,0,1) equals 3 gives k=-1,-1/2. Tangent planes: x+2y-2z+14=0 and 2x+2y-z-2=0. ## Answer **Sphere.** x² + y² + z² + 6x − 2z + 1 = 0 has centre C(−3, 0, 1) and radius r = √(9 + 0 + 1 − 1) = 3. **Line.** 3(16 − x) = 3z = 2y + 30 is the intersection of the planes x + z − 16 = 0 (from 3(16 − x) = 3z) and 3z − 2y − 30 = 0. #### Planes through the line ```text (x + z − 16) + k(−2y + 3z − 30) = 0 x − 2k y + (1 + 3k) z − (16 + 30k) = 0 ``` #### Tangent condition The distance from C to the plane must equal the radius 3. For the plane with coefficients A = 1, B = −2k, C₁ = 1 + 3k and D = −(16 + 30k): ```text A(−3) + B(0) + C₁(1) + D = −3 + 1 + 3k − 16 − 30k = −18 − 27k ``` Condition: (−18 − 27k)² = 9 [1 + 4k² + (1 + 3k)²] ```text 324 + 972k + 729k² = 9 (2 + 6k + 13k²) = 18 + 54k + 117k² 612k² + 918k + 306 = 0 2k² + 3k + 1 = 0 (2k + 1)(k + 1) = 0 → k = −1/2 or k = −1 ``` #### The planes - k = −1: x + 2y − 2z + 14 = 0, i.e. **x + 2y − 2z + 14 = 0** - k = −1/2: x + y − z/2 − 1 = 0, i.e. **2x + 2y − z − 2 = 0** Check: the point (16, −15, 0) lies on the line and on both planes (16 − 30 + 14 = 0; 32 − 30 − 2 = 0), and the distance from C to each plane is 9/3 = 3 = radius. **Answer: two tangent planes, x + 2y − 2z + 14 = 0 and 2x + 2y − z − 2 = 0.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the centre and radius of x^2+y^2+z^2+6x-2z+1 = 0? A. Centre (-3, 0, 1), radius 3 B. Centre (3, 0, -1), radius 3 C. Centre (-3, 0, 1), radius 9 D. Centre (-6, 0, 2), radius 6 Answer: A. u = 3, v = 0, w = -1, so centre (-3, 0, 1) and radius sqrt(9 + 0 + 1 - 1) = 3. (Key point: Sphere centre and radius) 2. How are planes through the line written? A. (x + z - 16) x k(3z - 2y - 30) = 0 B. (x + z - 16) + k(3z - 2y - 30) = 0 C. x + z - 16 + k = 0 D. k(x + z - 16) = 3z - 2y - 30 + k Answer: B. Any plane through the intersection of two planes is plane1 + k plane2 = 0. (Key point: Family of planes) 3. What condition makes the plane tangent? A. The plane passes through the centre B. The plane is parallel to the line C. The distance from the sphere's centre to the plane equals the radius 3 D. The distance from the origin to the plane equals 3 Answer: C. A tangent plane is at distance equal to the radius from the centre. (Key point: Tangency condition) 4. Solving 2k^2 + 3k + 1 = 0 gives how many tangent planes and which? A. Two: x + 2y - 2z + 14 = 0 and 2x + 2y - z - 2 = 0 B. One: x + 2y - 2z + 14 = 0 C. Two: x + 2y - 2z - 14 = 0 and 2x + 2y - z + 2 = 0 D. Two: x - 2y + 2z + 14 = 0 and 2x - 2y + z - 2 = 0 Answer: A. k = -1 and k = -1/2 give two tangent planes. 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