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 condition that holds true given the sum of the squares of the distances from the point (x, y, z) to the points (a, 0, 0) and (-a, 0, 0) equals \(2c^2\).

### Step-by-Step Solution:

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

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

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

4. **Add the Squares of the Distances**:
According to the problem, the sum of the squares of the distances is:
\[ d_1^2 + d_2^2 = 2c^2 \]
Substituting the squared distances:
\[ (x - a)^2 + y^2 + z^2 + (x + a)^2 + y^2 + z^2 = 2c^2 \]

5. **Expand the Squares**:
Expanding both squared terms:
\[ (x^2 - 2ax + a^2) + y^2 + z^2 + (x^2 + 2ax + a^2) + y^2 + z^2 = 2c^2 \]
Combining like terms:
\[ 2x^2 + 2y^2 + 2z^2 + 2a^2 = 2c^2 \]

6. **Simplify the Equation**:
Dividing the entire equation by 2:
\[ x^2 + y^2 + z^2 + a^2 = c^2 \]
Rearranging gives:
\[ x^2 + a^2 = c^2 - y^2 - z^2 \]

7. **Final Condition**:
Thus, the final condition we derived is:
\[ x^2 + a^2 = c^2 - y^2 - z^2 \]

### Conclusion:
The correct condition derived from the problem is:
\[ x^2 + a^2 = c^2 - y^2 - z^2 \]