1. **Problem Statement:** We have a square with vertices at $(-1,1)$, $(3,1)$, $(3,-3)$, and $(-1,-3)$ in the $xy$-plane. A circle is tangent to all four sides of this square. We need to find the equation of this circle. 2. **Understanding the Square:** The square's sides are vertical lines $x = -1$ and $x = 3$, and horizontal lines $y = 1$ and $y = -3$. 3. **Center of the Circle:** Since the circle is tangent to all four sides, it must be centered at the midpoint of the square. The midpoint (center) is the average of the $x$-coordinates and $y$-coordinates of opposite vertices: $$x_c = \frac{-1 + 3}{2} = 1, \quad y_c = \frac{1 + (-3)}{2} = -1$$ So, the center is at $(1, -1)$. 4. **Radius of the Circle:** The radius is the distance from the center to any side. The distance from $(1,-1)$ to the vertical side $x = 3$ is: $$r = |3 - 1| = 2$$ Similarly, the distance to the horizontal side $y = 1$ is: $$r = |1 - (-1)| = 2$$ This confirms the radius is $2$. 5. **Equation of the Circle:** The general equation of a circle with center $(h,k)$ and radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ Substituting $h=1$, $k=-1$, and $r=2$: $$ (x - 1)^2 + (y + 1)^2 = 4 $$ 6. **Answer:** The equation of the circle tangent to all four sides of the square is: $$\boxed{(x - 1)^2 + (y + 1)^2 = 4}$$