The sum of squares of distance of point P from (0,0) (0,1) (1,0),(1,1) = 18 then locus is at circle of diameter then find `d^2`

Answer

To solve the problem, we need to find the locus of point \( P(x, y) \) such that the sum of the squares of its distances from the points \( (0,0) \), \( (0,1) \), \( (1,0) \), and \( (1,1) \) equals 18.

### Step-by-Step Solution:

1. **Calculate the distances from point \( P \) to each of the four points:**
- Distance from \( P \) to \( (0,0) \):
\[ d_1 = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \]
- Distance from \( P \) to \( (0,1) \):
\[ d_2 = \sqrt{(x - 0)^2 + (y - 1)^2} = \sqrt{x^2 + (y - 1)^2} \]
- Distance from \( P \) to \( (1,0) \):
\[ d_3 = \sqrt{(x - 1)^2 + (y - 0)^2} = \sqrt{(x - 1)^2 + y^2} \]
- Distance from \( P \) to \( (1,1) \):
\[ d_4 = \sqrt{(x - 1)^2 + (y - 1)^2} = \sqrt{(x - 1)^2 + (y - 1)^2} \]

2. **Set up the equation for the sum of squares of distances:**
\[ d_1^2 + d_2^2 + d_3^2 + d_4^2 = 18 \]
Substituting the distances:
\[ (x^2 + y^2) + (x^2 + (y - 1)^2) + ((x - 1)^2 + y^2) + ((x - 1)^2 + (y - 1)^2) = 18 \]

3. **Expand each term:**
- \( d_2^2 = x^2 + (y^2 - 2y + 1) = x^2 + y^2 - 2y + 1 \)
- \( d_3^2 = ((x^2 - 2x + 1) + y^2) = x^2 - 2x + 1 + y^2 \)
- \( d_4^2 = ((x^2 - 2x + 1) + (y^2 - 2y + 1)) = x^2 - 2x + 1 + y^2 - 2y + 1 \)

4. **Combine all the terms:**
\[ x^2 + y^2 + (x^2 + y^2 - 2y + 1) + (x^2 - 2x + 1 + y^2) + (x^2 - 2x + 1 + y^2 - 2y + 1) = 18 \]
Simplifying gives:
\[ 4x^2 + 4y^2 - 4x - 4y + 4 = 18 \]

5. **Rearranging the equation:**
\[ 4x^2 + 4y^2 - 4x - 4y - 14 = 0 \]
Dividing through by 4:
\[ x^2 + y^2 - x - y - \frac{14}{4} = 0 \]
Which simplifies to:
\[ x^2 + y^2 - x - y - \frac{7}{2} = 0 \]

6. **Complete the square:**
\[ (x^2 - x) + (y^2 - y) = \frac{7}{2} \]
Completing the square for \( x \) and \( y \):
\[ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} = \frac{7}{2} \]
Thus:
\[ \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{7}{2} + \frac{1}{2} = 4 \]

7. **Identify the radius and diameter:**
The equation represents a circle with center \( \left(\frac{1}{2}, \frac{1}{2}\right) \) and radius \( r = 2 \). Therefore, the diameter \( d = 2r = 4 \).

8. **Find \( d^2 \):**
\[ d^2 = 4^2 = 16 \]

### Final Answer:
\[ d^2 = 16 \]