8. $[7]$ Two circles with radius one are drawn in the coordinate plane, one with center $(0,1)$ and the other with center $(2, y)$, for some real number $y$ between 0 and 1 . A third circle is drawn so as to be tangent to both of the other two circles as well as the $x$ axis. What is the smallest possible radius for this third circle?
Answer: $3-2 \sqrt{2}$ Suppose that the smaller circle has radius $r$. Call the three circles (in order from left to right) $O_{1}, O_{2}$, and $O_{3}$. The distance between the centers of $O_{1}$ and $O_{2}$ is $1+r$, and the distance in their $y$-coordinates is $1-r$. Therefore, by the Pythagorean theorem, the difference in $x$-coordinates
is $\sqrt{(1+r)^{2}-(1-r)^{2}}=2 \sqrt{r}$, which means that $O_{2}$ has a center at $(2 \sqrt{r}, r)$. But $O_{2}$ is also tangent to $\mathrm{O}_{3}$, which means that the difference in $x$-coordinate from the right-most point of $\mathrm{O}_{2}$ to the center of $\mathrm{O}_{3}$ is at most 1 . Therefore, the center of $O_{3}$ has an $x$-coordinate of at most $2 \sqrt{r}+r+1$, meaning that $2 \sqrt{r}+r+1 \leq 2$. We can use the quadratic formula to see that this implies that $\sqrt{r} \leq \sqrt{2}-1$, so $r \leq 3-2 \sqrt{2}$. We can achieve equality by placing the center of $O_{3}$ at $(2, r)$ (which in this case is $(2,3-2 \sqrt{2}))$