The locus of the point, the sum of the squares of whose distances from n fixed points `A _(i) (x _(i), y _(i)),i=1,2,…,n` is equal t `k ^(2)` is a circle

Options

A. passing through the origin
B. with centre at the origin
C. with cnetre at the point of mean position of the given points
D. none of these

Answer

To solve the problem, we need to find the locus of a point \( P(h, k) \) such that the sum of the squares of its distances from \( n \) fixed points \( A_i(x_i, y_i) \) is equal to \( k^2 \).

### Step-by-Step Solution:

1. **Understanding the Distance Formula**:
The distance from point \( P(h, k) \) to a fixed point \( A_i(x_i, y_i) \) is given by:
\[ d_i = \sqrt{(h - x_i)^2 + (k - y_i)^2} \]
Therefore, the square of the distance is:
\[ d_i^2 = (h - x_i)^2 + (k - y_i)^2 \]

2. **Setting Up the Equation**:
We need to sum the squares of the distances from \( P(h, k) \) to all \( n \) fixed points and set this equal to \( k^2 \):
\[ \sum_{i=1}^{n} d_i^2 = k^2 \]
Substituting the expression for \( d_i^2 \):
\[ \sum_{i=1}^{n} \left( (h - x_i)^2 + (k - y_i)^2 \right) = k^2 \]

3. **Expanding the Equation**:
Expanding the left-hand side:
\[ \sum_{i=1}^{n} (h^2 - 2hx_i + x_i^2 + k^2 - 2ky_i + y_i^2) = k^2 \]
This simplifies to:
\[ n(h^2 + k^2) - 2h\sum_{i=1}^{n} x_i - 2k\sum_{i=1}^{n} y_i + \sum_{i=1}^{n} (x_i^2 + y_i^2) = k^2 \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ n(h^2 + k^2) - 2h\sum_{i=1}^{n} x_i - 2k\sum_{i=1}^{n} y_i + \sum_{i=1}^{n} (x_i^2 + y_i^2) - k^2 = 0 \]

5. **Identifying the Circle Equation**:
This is a quadratic equation in \( h \) and \( k \):
\[ nh^2 + nk^2 - 2h\sum_{i=1}^{n} x_i - 2k\sum_{i=1}^{n} y_i + \left(\sum_{i=1}^{n} (x_i^2 + y_i^2) - k^2\right) = 0 \]
This can be rearranged into the standard form of a circle:
\[ (h - \bar{x})^2 + (k - \bar{y})^2 = r^2 \]
where \( \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \) and \( \bar{y} = \frac{\sum_{i=1}^{n} y_i}{n} \) are the mean coordinates of the fixed points, and \( r \) is the radius determined by the constant terms.

6. **Conclusion**:
Thus, the locus of the point \( P(h, k) \) is a circle with center at \( \left( \frac{\sum_{i=1}^{n} x_i}{n}, \frac{\sum_{i=1}^{n} y_i}{n} \right) \).