A circle of radius $2$ has center at $(2,0)$. A circle of radius $1$ has center at $(5,0)$. A line is tangent to the two circles at points in the first quadrant. Which of the following is closest to the $y$-intercept of the line?
$\text{(A)}\ \sqrt{2}/4\qquad\text{(B)}\ 8/3\qquad\text{(C)}\ 1+\sqrt 3\qquad\text{(D)}\ 2\sqrt 2\qquad\text{(E)}\ 3$
The two circles are tangent to each other at the point $(4,0)$, since it is both $2$ units from $(2,0)$ and $1$ unit from $(5,0)$.
Label the x-intercept of the common tangent line $A$, and label the y-intercept of the common tangent $B$.  Triangle $\triangle OAB$ is a right triangle at the origin.
Label $D$ the point of tangency to the first, big circle, and label $E$ the point of tangency to the small circle.  $\angle D$ and $\angle E$ are both right angles as well.
Label the center of the big circle $F$, and the small circle $G$.
$\triangle DAF  \sim \triangle EAG \sim \triangle OAB$ because they each have one right angle, and also have a common angle $A$.
The y-intercept is the length $OB$.
We know that $DF = 2$ and $EG = 1$ because they are radii of circles.  From similarity, $\frac{DF}{EG} = \frac{FA}{GA}$.  Thus, $FA = 2GA$.
Looking at line $FGA$, we know that $FG + GA = FA$.
Thus, $FG + GA = 2GA$, meaning $FG = GA$.  Since $FG = 3$ because it's the distance of the centers of the circles, so $GA = 3$ as well.  This gives point $A$ as $(8,0)$.
Since $\triangle EAG$ is a right triangle with $GA = 3$ and $GE = 1$, we know that $AE = \sqrt{GA^2 - GE^2} = \sqrt{8}$
Thus, all triangles are $1:\sqrt{8}:3$ triangles.
$\triangle OAC$ has as its middle side $OA$, which is $8$.  Thus, all sides are scaled up by a factor of $\sqrt{8}$, and $OB$, the shortest side, is $\sqrt{8}$.  This means the y-intercept is $\sqrt{8} = 2\sqrt{2}$, and the answer is $\boxed{D}$.