Find the equation of the locus of point P, the sum of the square of whose distances from the
points A(0, 6, 0) and B(0, -6, 0) is 100.

Answer

To find the equation of the locus of point P such that the sum of the squares of its distances from points A(0, 6, 0) and B(0, -6, 0) is 100, we will follow these steps:

### Step 1: Define the distances
Let P be a point with coordinates (x, y, z). We need to calculate the distances from P to points A and B.

The distance from P to A is given by:
\[ AP = \sqrt{(x - 0)^2 + (y - 6)^2 + (z - 0)^2} = \sqrt{x^2 + (y - 6)^2 + z^2} \]

The distance from P to B is given by:
\[ PB = \sqrt{(x - 0)^2 + (y + 6)^2 + (z - 0)^2} = \sqrt{x^2 + (y + 6)^2 + z^2} \]

### Step 2: Square the distances
According to the problem, the sum of the squares of these distances equals 100:
\[ AP^2 + PB^2 = 100 \]

Substituting the expressions for AP and PB:
\[ (x^2 + (y - 6)^2 + z^2) + (x^2 + (y + 6)^2 + z^2) = 100 \]

### Step 3: Expand the squares
Now we expand the squared terms:
1. For \( (y - 6)^2 \):
\[ (y - 6)^2 = y^2 - 12y + 36 \]

2. For \( (y + 6)^2 \):
\[ (y + 6)^2 = y^2 + 12y + 36 \]

Now substituting these into the equation:
\[ x^2 + (y^2 - 12y + 36) + z^2 + x^2 + (y^2 + 12y + 36) + z^2 = 100 \]

### Step 4: Combine like terms
Combining the terms gives:
\[ 2x^2 + 2y^2 + 2z^2 + 72 = 100 \]

### Step 5: Simplify the equation
Now, we simplify this equation:
\[ 2x^2 + 2y^2 + 2z^2 = 100 - 72 \]
\[ 2x^2 + 2y^2 + 2z^2 = 28 \]

Dividing the entire equation by 2:
\[ x^2 + y^2 + z^2 = 14 \]

### Conclusion
The equation of the locus of point P is:
\[ x^2 + y^2 + z^2 = 14 \]

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