Suppose the transformation $T$ acts on points in the plane like this:
$$T(x, y) =  \left( \frac{x}{x^2 + y^2}, \frac{-y}{x^2 + y^2}\right).$$
Determine the area enclosed by the set of points of the form $T(x, y)$, where $(x, y)$ is a point on the edge of a length-$2$ square centered at the origin with sides parallel to the axes.
1. **Understanding the Transformation:**
   The given transformation \( T \) is defined as:
   \[
   T(x, y) = \left( \frac{x}{x^2 + y^2}, \frac{-y}{x^2 + y^2} \right).
   \]
   This transformation can be interpreted in terms of complex numbers. If we let \( z = x + iy \), then the transformation \( T \) can be written as:
   \[
   T(z) = \frac{1}{z}.
   \]
   This is known as the inversion transformation in the complex plane.

2. **Inversion in the Complex Plane:**
   The inversion transformation \( \phi \colon z \mapsto \frac{1}{\overline{z}} \) maps points inside the unit circle to points outside the unit circle and vice versa. For a point \( z = x + iy \), the transformation \( \phi \) is given by:
   \[
   \phi(z) = \frac{1}{\overline{z}} = \frac{1}{x - iy} = \frac{x + iy}{x^2 + y^2}.
   \]
   This matches the given transformation \( T \).

3. **Mapping the Square:**
   Consider the square centered at the origin with side length 2. The vertices of this square are \((\pm 1, \pm 1)\). The edges of the square are given by the lines:
   \[
   x = \pm 1, \quad -1 \leq y \leq 1 \quad \text{and} \quad y = \pm 1, \quad -1 \leq x \leq 1.
   \]

4. **Transforming the Edges:**
   We need to determine the image of these edges under the transformation \( T \).

   - For the edge \( x = 1 \):
     \[
     T(1, y) = \left( \frac{1}{1 + y^2}, \frac{-y}{1 + y^2} \right).
     \]
     As \( y \) ranges from \(-1\) to \(1\), the points trace out a curve.

   - For the edge \( x = -1 \):
     \[
     T(-1, y) = \left( \frac{-1}{1 + y^2}, \frac{-y}{1 + y^2} \right).
     \]
     As \( y \) ranges from \(-1\) to \(1\), the points trace out a curve.

   - For the edge \( y = 1 \):
     \[
     T(x, 1) = \left( \frac{x}{x^2 + 1}, \frac{-1}{x^2 + 1} \right).
     \]
     As \( x \) ranges from \(-1\) to \(1\), the points trace out a curve.

   - For the edge \( y = -1 \):
     \[
     T(x, -1) = \left( \frac{x}{x^2 + 1}, \frac{1}{x^2 + 1} \right).
     \]
     As \( x \) ranges from \(-1\) to \(1\), the points trace out a curve.

5. **Shape of the Transformed Region:**
   By symmetry and the nature of the inversion transformation, the image of the square under \( T \) is a circle. This is because the inversion transformation maps the boundary of the square to a circle.

6. **Area of the Transformed Region:**
   The area of the circle can be determined by noting that the inversion transformation preserves the area. The original square has an area of \( 4 \) (since it is a \( 2 \times 2 \) square). The transformed region, being a circle, must have the same area.

   The area of a circle is given by \( \pi r^2 \). Setting this equal to the area of the square, we get:
   \[
   \pi r^2 = 4 \implies r^2 = \frac{4}{\pi} \implies r = \sqrt{\frac{4}{\pi}}.
   \]

   Therefore, the radius of the circle is \( \sqrt{\frac{4}{\pi}} \), and the area of the circle is \( 4 \).

\(\blacksquare\)

The final answer is \( \boxed{ 4 } \).