1. **Problem Statement:** We have a square with vertices at $(-1,1)$, $(3,1)$, $(3,-3)$, and $(-1,-3)$, and circles defined by equations such as $(x+1)^2+(y-1)^2=4$, $(x-1)^2+(y+1)^2=4$, $(x+1)^2+(y-1)^2=9$, $(x-1)^2+(y+1)^2=9$, and $(x+2)^2+(y-2)^2=9$. We want to understand the relationship between the square and the circles, especially the green circle tangent to each side of the square. 2. **Square Properties:** The square's side length is the distance between $(-1,1)$ and $(3,1)$, which is $3 - (-1) = 4$. The square is axis-aligned with sides parallel to the axes. 3. **Circle Equations:** The general form of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. 4. **Check the circles:** - $(x+1)^2+(y-1)^2=4$ has center $(-1,1)$ and radius $2$. - $(x-1)^2+(y+1)^2=4$ has center $(1,-1)$ and radius $2$. - $(x+1)^2+(y-1)^2=9$ has center $(-1,1)$ and radius $3$. - $(x-1)^2+(y+1)^2=9$ has center $(1,-1)$ and radius $3$. - $(x+2)^2+(y-2)^2=9$ has center $(-2,2)$ and radius $3$. 5. **Green Circle Tangent to Square Sides:** The green circle tangent to each side of the square must be inscribed in the square. The square's center is at the midpoint of the diagonal: $$\left(\frac{-1+3}{2}, \frac{1+(-3)}{2}\right) = (1,-1).$$ 6. **Radius of Inscribed Circle:** The inscribed circle radius equals half the side length of the square, so $$r = \frac{4}{2} = 2.$$ The circle centered at $(1,-1)$ with radius $2$ is given by: $$ (x-1)^2 + (y+1)^2 = 4. $$ 7. **Verification:** This circle touches all four sides of the square exactly once, confirming it is tangent to each side. **Final answer:** The green circle tangent to each side of the square is centered at $(1,-1)$ with radius $2$, described by the equation: $$ (x-1)^2 + (y+1)^2 = 4. $$