Find the locus of a point which moves in such a way that the sum of its distances from the points `(a, 0, 0) and (a, 0, 0)` is constant.

Explanation

To find the locus of a point \( P(x, y, z) \) such that the sum of its distances from the points \( A(a, 0, 0) \) and \( B(a, 0, 0) \) is constant, we can follow these steps:

### Step 1: Define the distances
The distance from point \( P(x, y, z) \) to point \( A(a, 0, 0) \) is given by:
\[ d_1 = \sqrt{(x - a)^2 + y^2 + z^2} \]
Since points \( A \) and \( B \) are the same, the distance from \( P \) to \( B \) is the same:
\[ d_2 = \sqrt{(x - a)^2 + y^2 + z^2} \]

### Step 2: Set up the equation for the sum of distances
The sum of the distances from point \( P \) to points \( A \) and \( B \) is constant, say \( k \):
\[ d_1 + d_2 = k \]
Since \( d_1 = d_2 \), we can rewrite this as:
\[ 2d_1 = k \]
Thus, we have:
\[ d_1 = \frac{k}{2} \]

### Step 3: Substitute the expression for distance
Substituting the expression for \( d_1 \):
\[ \sqrt{(x - a)^2 + y^2 + z^2} = \frac{k}{2} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square root gives:
\[ (x - a)^2 + y^2 + z^2 = \left(\frac{k}{2}\right)^2 \]
This simplifies to:
\[ (x - a)^2 + y^2 + z^2 = \frac{k^2}{4} \]

### Step 5: Rearranging the equation
This equation represents a sphere centered at \( (a, 0, 0) \) with radius \( \frac{k}{2} \). Therefore, the locus of point \( P \) is:
\[ (x - a)^2 + y^2 + z^2 = \frac{k^2}{4} \]

### Final Answer
The locus of the point \( P \) is a sphere centered at \( (a, 0, 0) \) with radius \( \frac{k}{2} \).

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