1. **Problem Statement:** We have a square of side length 4 units. - Semicircle A with radius 2 units is drawn on the left side. - Quarter circle B with radius 2 units is drawn at the bottom-right corner. - Circle C touches semicircle A and quarter circle B externally and is tangent internally to the right side of the square. We need to find the radius $r$ of circle C. 2. **Setup and Known Data:** - Square side length $=4$. - Radius of semicircle A, $R_A=2$. - Radius of quarter circle B, $R_B=2$. - Circle C radius $=r$ (unknown). 3. **Positions and Centers:** - Center of semicircle A is at $(0,2)$ (middle of left side). - Center of quarter circle B is at $(4,0)$ (bottom-right corner). - Circle C is tangent to the right side of the square, so its center lies at $(4 - r, y)$ for some $y$. 4. **Tangency Conditions:** - Circle C touches semicircle A externally: $$\text{distance between centers} = R_A + r$$ $$\sqrt{(4 - r - 0)^2 + (y - 2)^2} = 2 + r$$ - Circle C touches quarter circle B externally: $$\text{distance between centers} = R_B + r$$ $$\sqrt{(4 - r - 4)^2 + (y - 0)^2} = 2 + r$$ - Circle C is tangent internally to the right side of the square, so its center is at $x = 4 - r$. 5. **Simplify the equations:** - From the second tangency: $$\sqrt{(0)^2 + (y)^2} = 2 + r \implies y = 2 + r$$ - Substitute $y = 2 + r$ into the first tangency: $$\sqrt{(4 - r)^2 + ((2 + r) - 2)^2} = 2 + r$$ $$\sqrt{(4 - r)^2 + r^2} = 2 + r$$ 6. **Square both sides:** $$(4 - r)^2 + r^2 = (2 + r)^2$$ $$16 - 8r + r^2 + r^2 = 4 + 4r + r^2$$ $$16 - 8r + 2r^2 = 4 + 4r + r^2$$ 7. **Bring all terms to one side:** $$16 - 8r + 2r^2 - 4 - 4r - r^2 = 0$$ $$12 - 12r + r^2 = 0$$ 8. **Rewrite quadratic:** $$r^2 - 12r + 12 = 0$$ 9. **Solve quadratic equation:** $$r = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 1 \times 12}}{2} = \frac{12 \pm \sqrt{144 - 48}}{2} = \frac{12 \pm \sqrt{96}}{2}$$ $$= \frac{12 \pm 4\sqrt{6}}{2} = 6 \pm 2\sqrt{6}$$ 10. **Choose the valid radius:** - Radius must be positive and less than 4 (since circle C fits inside the square). - $6 + 2\sqrt{6} > 4$ (too large), discard. - $6 - 2\sqrt{6} \approx 6 - 4.9 = 1.1$ (valid). **Final answer:** $$\boxed{r = 6 - 2\sqrt{6}}$$