If the sum of the squares of the distances of the
point (x, y, z) from the points (a, 0, 0) and (-a, 0, 0)
is `2c^(2)`, then which are of the following is
correct?

Options

A. `x^(2)-a^(2)=c^(2)-y^(2)-z^(2)`
B. `x^(2)+a^(2)=c^(2)+y^(2)+z^(2)`
C. `x^(2)+a^(2)=2c^(2)-y^(2)-z^(2)`
D. `x^(2)+a^(2)=c^(2)-y^(2)-z^(2)`

Answer

To solve the problem, we need to find the sum of the squares of the distances from the point \( (x, y, z) \) to the points \( (a, 0, 0) \) and \( (-a, 0, 0) \), and set it equal to \( 2c^2 \).

### Step-by-Step Solution:

1. **Calculate the distance from \( (x, y, z) \) to \( (a, 0, 0) \)**:
\[ d_1 = \sqrt{(x - a)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{(x - a)^2 + y^2 + z^2} \]

2. **Calculate the distance from \( (x, y, z) \) to \( (-a, 0, 0) \)**:
\[ d_2 = \sqrt{(x + a)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{(x + a)^2 + y^2 + z^2} \]

3. **Square the distances**:
\[ d_1^2 = (x - a)^2 + y^2 + z^2 \]
\[ d_2^2 = (x + a)^2 + y^2 + z^2 \]

4. **Sum the squares of the distances**:
\[ d_1^2 + d_2^2 = [(x - a)^2 + y^2 + z^2] + [(x + a)^2 + y^2 + z^2] \]

5. **Expand the squares**:
\[ (x - a)^2 = x^2 - 2ax + a^2 \]
\[ (x + a)^2 = x^2 + 2ax + a^2 \]
Therefore,
\[ d_1^2 + d_2^2 = [x^2 - 2ax + a^2 + y^2 + z^2] + [x^2 + 2ax + a^2 + y^2 + z^2] \]

6. **Combine like terms**:
\[ d_1^2 + d_2^2 = 2x^2 + 2a^2 + 2y^2 + 2z^2 \]
\[ = 2(x^2 + a^2 + y^2 + z^2) \]

7. **Set the sum equal to \( 2c^2 \)**:
\[ 2(x^2 + a^2 + y^2 + z^2) = 2c^2 \]

8. **Divide both sides by 2**:
\[ x^2 + a^2 + y^2 + z^2 = c^2 \]

### Conclusion:
The equation \( x^2 + a^2 + y^2 + z^2 = c^2 \) represents a sphere in three-dimensional space.

### Final Answer:
The correct option is \( x^2 + a^2 + y^2 + z^2 = c^2 \).