Q213 / 270 · NDA 2026 (Phase 1) · Mathematics
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 $S$ ?
Question: What is the radius of the sphere passing through origin and concentric with the sphere $S$ ?
- A$\dfrac{7}{2}$
- B$5$
- Correct answerC$7$
- DCannot be determined due to insufficient data
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.
$R = \sqrt{(2-0)^2 + (3-0)^2 + (6-0)^2}$
$R = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$
Answer: $7$
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.
$R = \sqrt{(2-0)^2 + (3-0)^2 + (6-0)^2}$
$R = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$
Answer: $7$
Hence, the correct answer is option: 'C'.