--- title: "Plane touches sphere, find point of contact" subject: "Mathematics" chapter: 3 key: "semester_2/mathematics:plane-touches-sphere-find-point" source: https://www.bctengineer.com/q/semester_2/mathematics/plane-touches-sphere-find-point site: BCT Engineering --- # Plane touches sphere, find point of contact Mathematics · Chapter 3: Three Dimensional Solid Geometry - Asked in: 2071 Magh, 5 marks - Video: [Problems on the Point of Contact Where the Plane Touches the Sphere / (Unit-3) / Degree 2nd Sem /...](https://www.youtube.com/watch?v=NMMLJQnFJuM) — DMT LOGICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/mathematics/plane-touches-sphere-find-point - Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3 ## Question Prove that the plane 2x−y+2z−14 = 0 touches the sphere x²+y²+z²−4x+2y−4 = 0. Find the point of contact. ## Short answer Sphere centre (2,-1,0), radius 3. Distance from centre to plane 2x-y+2z-14=0 is 3, equal to radius, so it touches. Point of contact, foot of perpendicular: (4,-2,2). ## Answer **Sphere.** x² + y² + z² − 4x + 2y − 4 = 0. Comparing with x² + y² + z² + 2ux + 2vy + 2wz + d = 0: u = −2, v = 1, w = 0, d = −4. ```text Centre C = (2, −1, 0) Radius r = √(4 + 1 + 0 + 4) = 3 ``` #### Proof of tangency A plane touches a sphere when the perpendicular distance from the centre to the plane equals the radius. Plane: 2x − y + 2z − 14 = 0, with |normal| = √(4 + 1 + 4) = 3. ```text d = |2(2) − (−1) + 2(0) − 14| / 3 = |4 + 1 − 14| / 3 = 9/3 = 3 ``` This equals the radius 3, so the plane touches the sphere. ∎ #### Point of contact It is the foot of the perpendicular from C to the plane. The perpendicular from C has direction (2, −1, 2): (2 + 2t, −1 − t, 2t). Put on the plane: ```text 2(2 + 2t) − (−1 − t) + 2(2t) − 14 = 0 4 + 4t + 1 + t + 4t − 14 = 0 9t = 9 → t = 1 ``` Point of contact: (4, −2, 2). Check: on the sphere: 16 + 4 + 4 − 16 − 4 − 4 = 0 ✓. On the plane: 8 + 2 + 4 − 14 = 0 ✓. **Answer: the point of contact is (4, −2, 2).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When does a plane touch a sphere? A. The distance from the centre to the plane is zero B. The plane passes through the origin C. The distance from the centre to the plane equals the radius D. The distance from the origin to the plane equals the radius Answer: C. A tangent plane is at distance r from the centre. (Key point: Tangency condition) 2. For x^2+y^2+z^2-4x+2y-4 = 0, what are the centre and radius? A. (-2, 1, 0), 3 B. (2, -1, 0), 3 C. (2, -1, 0), 9 D. (2, -1, 0), sqrt(5) Answer: B. u = -2, v = 1, w = 0, d = -4 give centre (2, -1, 0) and r = sqrt(4+1+0+4) = 3. (Key point: Centre and radius) 3. How is the point of contact found? A. As the foot of the perpendicular from the centre, along the plane's normal B. As the intersection of the plane with the x-axis C. As the midpoint of the centre and the origin D. By setting z = 0 in the plane Answer: A. The radius to the touching point is perpendicular to the plane. (Key point: Point of contact) 4. What is the point of contact? A. (2, -1, 2) B. (4, -2, 2) C. (4, -2, 0) D. (0, 0, 7) Answer: B. (2+2t, -1-t, 2t) with t = 1 gives (4, -2, 2). 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