Question
Passage: The equation of the sphere S is x^2+y^2+z^2-4x-6y-12z+k=0. Question: What is the radius of the sphere passing through origin and concentric with the sphere S ?
NDA
Mathematics
Three Dimensional Geometry
2026
NDA 2026 (Phase 1)
NDA Mathematics Question (2026) — Solution
Options
- A. 7 2
- B. 5
- C. 7
- D. Cannot be determined due to insufficient data
Answer
C. 7
Step-by-step solution
The equation of the given sphere is x^2+y^2+z^2-4x-6y-12z+k=0. The center of the sphere is (2, 3, 6). The required sphere is concentric with the given sphere, so its center is also (2, 3, 6). Since the required sphere passes through the origin (0, 0, 0), its radius is the distance between the center and the origin. R = (2-0)^2 + (3-0)^2 + (6-0)^2 R = 4 + 9 + 36 = 49 = 7 Answer: 7
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