1. **Stating the problem:** We have a square of side length 4 meters. Inside the square, a vertical line divides it into two equal parts (each 2 meters wide). A horizontal line divides the right half at mid-height, but this does not affect the lower-left area. A circle is placed in the lower-left area, tangent to the left and bottom sides of the square, and also tangent to the central vertical line. We need to find the radius $r$ of this circle. 2. **Understanding the setup:** - The square side length is 4 m. - The vertical line at midpoint divides the square into two rectangles of width 2 m each. - The circle is tangent to the left side (x=0), bottom side (y=0), and the vertical line at x=2. 3. **Using the tangent conditions:** - Since the circle is tangent to the left side (x=0) and bottom side (y=0), its center must be at $(r, r)$ because the radius is the distance from the center to these sides. - The circle is also tangent to the vertical line at $x=2$. 4. **Formulating the tangent condition to the vertical line:** - The distance from the center $(r, r)$ to the vertical line $x=2$ must be equal to the radius $r$. - The distance from $(r, r)$ to $x=2$ is $2 - r$. - So, $2 - r = r$. 5. **Solving for $r$:** $$ 2 - r = r \\ 2 = 2r \\ r = \frac{2}{2} = 1 $$ 6. **Final answer:** The radius of the circle is $\boxed{1}$ meter.