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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2026 · Mathematics · Three Dimensional Geometry

Passage: The equation of the sphere S is x^2+y^2+z^2-4x-6y-12z+k=0...

Passage: The equation of the sphere SS is x2+y2+z2−4x−6y−12z+k=0x^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 SS ?

Options

  1. A72\dfrac{7}{2}
  2. B55
  3. C77
  4. DCannot be determined due to insufficient data

Correct answer

C. 77

Step-by-step solution

The equation of the given sphere is x2+y2+z2−4x−6y−12z+k=0x^2+y^2+z^2-4x-6y-12z+k=0.

The center of the sphere is (2,3,6)(2, 3, 6).

The required sphere is concentric with the given sphere, so its center is also (2,3,6)(2, 3, 6).

Since the required sphere passes through the origin (0,0,0)(0, 0, 0), its radius is the distance between the center and the origin.

R=(2−0)2+(3−0)2+(6−0)2R = \sqrt{(2-0)^2 + (3-0)^2 + (6-0)^2}

R=4+9+36=49=7R = \sqrt{4 + 9 + 36} = \sqrt{49} = 7

Answer: 77
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