A sphere of constant radius `2k` passes through the origin and meets the axes in `A ,B ,a n dCdot` The locus of a centroid of the tetrahedron `O A B C` is a. `x^2+y^2+z^2=4k^2` b. `x^2+y^2+z^2=k^2` c. `2(x^2+y^2+z)^2=k^2` d. none of these

Options

A. `x^(2)+y^(2)+z^(2)=k^(2)`
B. `x^(2)+y^(2)+z^(2)=k^(2)`
C. `2(k^(2)+y^(2)+z)^(2)=k^(2)`
D. none of these

Explanation

To solve the problem, we need to find the locus of the centroid of the tetrahedron formed by the origin \( O \) and the points \( A \), \( B \), and \( C \) where the sphere of radius \( 2k \) intersects the axes.

### Step-by-Step Solution:

1. **Identify the Coordinates of Points A, B, and C**:
- The sphere intersects the x-axis at point \( A \), the y-axis at point \( B \), and the z-axis at point \( C \).
- Let the coordinates of these points be:
- \( A = (a, 0, 0) \)
- \( B = (0, b, 0) \)
- \( C = (0, 0, c) \)

2. **Equation of the Sphere**:
- The general equation of a sphere centered at the origin with radius \( r \) is given by:
\[ x^2 + y^2 + z^2 = r^2 \]
- Given that the sphere has a radius of \( 2k \), we have:
\[ x^2 + y^2 + z^2 = (2k)^2 = 4k^2 \]

3. **Using the Sphere's Equation**:
- Since the sphere passes through points \( A \), \( B \), and \( C \), we can write the equation of the sphere in terms of \( a \), \( b \), and \( c \):
\[ x^2 + y^2 + z^2 - ax - by - cz = 0 \]
- From the sphere's equation, we know:
\[ a^2 + b^2 + c^2 = 4(2k)^2 = 16k^2 \]
- This can be labeled as Equation (1).

4. **Finding the Centroid of Tetrahedron OABC**:
- The coordinates of the centroid \( G \) of tetrahedron \( OABC \) are given by:
\[ G\left(\frac{0 + a + 0 + 0}{4}, \frac{0 + 0 + b + 0}{4}, \frac{0 + 0 + 0 + c}{4}\right) = \left(\frac{a}{4}, \frac{b}{4}, \frac{c}{4}\right) \]
- Let \( x = \frac{a}{4} \), \( y = \frac{b}{4} \), \( z = \frac{c}{4} \).
- Therefore, we can express \( a \), \( b \), and \( c \) in terms of \( x \), \( y \), and \( z \):
\[ a = 4x, \quad b = 4y, \quad c = 4z \]
- This can be labeled as Equation (2).

5. **Substituting into Equation (1)**:
- Substitute \( a \), \( b \), and \( c \) from Equation (2) into Equation (1):
\[ (4x)^2 + (4y)^2 + (4z)^2 = 16k^2 \]
- Simplifying this gives:
\[ 16(x^2 + y^2 + z^2) = 16k^2 \]
- Dividing both sides by 16:
\[ x^2 + y^2 + z^2 = k^2 \]

6. **Conclusion**:
- The locus of the centroid of the tetrahedron \( OABC \) is:
\[ x^2 + y^2 + z^2 = k^2 \]
- Therefore, the correct answer is option **b**: \( x^2 + y^2 + z^2 = k^2 \).