The extremities of a diagonal of a rectangle are (0.0) and (4, 4). The locus of the extremities of the other diagonal is equal to

Options

A. `x^(2) +y^(2) - 4x - 4y = 0`
B. `x^(2)+y^(2)+4x +4y - 4 = 0`
C. `x^(2)+y^(2)+4x +4y +4 = 0`
D. `x^(2)+y^(2) - 4x - 4y - 4 = 0`

Explanation

To find the locus of the extremities of the other diagonal of the rectangle given the extremities of one diagonal at points (0, 0) and (4, 4), we can follow these steps:

### Step 1: Understand the Geometry of the Rectangle
The rectangle has vertices A(0, 0) and C(4, 4). Let the other two vertices be B(h, k) and D(4 + h, k) such that the diagonals AC and BD intersect at the midpoint of both diagonals.

### Step 2: Find the Midpoint of Diagonal AC
The midpoint M of diagonal AC can be calculated as:
\[ M = \left( \frac{0 + 4}{2}, \frac{0 + 4}{2} \right) = (2, 2) \]

### Step 3: Use the Property of Perpendicularity
Since AB is perpendicular to BC, we can use the slopes to find a relationship between h and k. The slope of AB is:
\[ \text{slope of AB} = \frac{k - 0}{h - 0} = \frac{k}{h} \]
The slope of BC is:
\[ \text{slope of BC} = \frac{k - 4}{h - 4} \]
Since AB is perpendicular to BC, we have:
\[ \frac{k}{h} \cdot \frac{k - 4}{h - 4} = -1 \]

### Step 4: Cross-Multiply and Rearrange
Cross-multiplying gives:
\[ k(k - 4) = -h(h - 4) \]
Expanding both sides:
\[ k^2 - 4k = -h^2 + 4h \]
Rearranging gives:
\[ k^2 + h^2 - 4h - 4k = 0 \]

### Step 5: Express in Terms of x and y
Let \( x = h \) and \( y = k \). Then we can rewrite the equation as:
\[ x^2 + y^2 - 4x - 4y = 0 \]

### Step 6: Identify the Locus
The equation \( x^2 + y^2 - 4x - 4y = 0 \) represents the locus of the extremities of the other diagonal of the rectangle.

### Conclusion
The locus of the extremities of the other diagonal is given by:
\[ x^2 + y^2 - 4x - 4y = 0 \]

### Final Answer
The correct option is:
**Option 1: \( x^2 + y^2 - 4x - 4y = 0 \)**

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