The locus of a point which moves such that the sum of the squares of the distances from the three vertices of a triangle is constant, is a circle whose centre is at the:

Options

A. Incentre of the triangle
B. Centroid of the triangle
C. Orthocentre of the triangle
D. Circumcentre of the triangle

Explanation

To solve the problem, we need to find the locus of a point \( P(x, y) \) such that the sum of the squares of the distances from the point \( P \) to the three vertices of a triangle \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is constant.

### Step-by-Step Solution:

1. **Define the Distances**:
The distance from point \( P \) to vertex \( A \) is given by:
\[ AP = \sqrt{(x - x_1)^2 + (y - y_1)^2} \]
Similarly, the distances to vertices \( B \) and \( C \) are:
\[ BP = \sqrt{(x - x_2)^2 + (y - y_2)^2} \]
\[ CP = \sqrt{(x - x_3)^2 + (y - y_3)^2} \]

2. **Square the Distances**:
We square the distances to avoid dealing with square roots:
\[ AP^2 = (x - x_1)^2 + (y - y_1)^2 \]
\[ BP^2 = (x - x_2)^2 + (y - y_2)^2 \]
\[ CP^2 = (x - x_3)^2 + (y - y_3)^2 \]

3. **Set Up the Equation**:
According to the problem, the sum of these squared distances is constant:
\[ AP^2 + BP^2 + CP^2 = k \quad (k \text{ is a constant}) \]

4. **Expand the Squared Distances**:
Expanding each squared distance, we have:
\[ (x - x_1)^2 + (y - y_1)^2 + (x - x_2)^2 + (y - y_2)^2 + (x - x_3)^2 + (y - y_3)^2 \]
Expanding each term:
\[ = (x^2 - 2xx_1 + x_1^2 + y^2 - 2yy_1 + y_1^2) + (x^2 - 2xx_2 + x_2^2 + y^2 - 2yy_2 + y_2^2) + (x^2 - 2xx_3 + x_3^2 + y^2 - 2yy_3 + y_3^2) \]

5. **Combine Like Terms**:
Combine all the terms:
\[ 3x^2 + 3y^2 - 2x(x_1 + x_2 + x_3) - 2y(y_1 + y_2 + y_3) + (x_1^2 + y_1^2 + x_2^2 + y_2^2 + x_3^2 + y_3^2) = k \]

6. **Rearrange the Equation**:
Rearranging gives us:
\[ 3x^2 + 3y^2 - 2x\left(\frac{x_1 + x_2 + x_3}{3}\right) - 2y\left(\frac{y_1 + y_2 + y_3}{3}\right) = k - \text{constant terms} \]

7. **Identify the Circle**:
This equation represents a circle. The center of the circle can be identified as:
\[ \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
This point is the centroid of triangle \( ABC \).

### Conclusion:
The locus of the point \( P \) is a circle whose center is at the centroid of triangle \( ABC \).