$26 \cdot 31$ One circle has its center at $(2,0)$ and radius 2, another circle has its center at $(5,0)$ and radius 1. A line is tangent to both circles in the first quadrant. The $y$-intercept of this line is closest to
(A) $\frac{\sqrt{2}}{4}$.
(B) $\frac{8}{3}$.
(C) $1+\sqrt{3}$.
(D) $2 \sqrt{2}$.
(E) 3.
(47th American High School Mathematics Examination, 1996)
[Solution] As shown in the figure, it is easy to prove $\triangle C B O_{2} \sim s$
$$
\begin{aligned}
\triangle C A O_{1} & \sim \triangle O_{2} E O_{1} \backsim \triangle C O D . \\
\text { By } & \frac{O_{1} A}{O_{1} C}=\frac{O_{2} B}{O_{2} C} \\
= & \frac{O_{1} A-O_{2} B}{O_{1} C-O_{2} C} \\
\therefore \quad & O_{2} C=O_{2} B \cdot \frac{O_{1} C-O_{2} C}{O_{1} A-O_{2} B}=1 \cdot \frac{5-2}{2-1}=3, \\
\text { and } & B=\sqrt{8} .
\end{aligned}
$$

At the same time, $\frac{O D}{O_{2} B}=\frac{O C}{C B}$, and $O C=O O_{1}+O_{2} C=8$,
$$
\therefore \quad O D=\frac{O_{2} B \cdot O C}{C B}=\frac{1 \cdot 8}{\sqrt{8}}=\sqrt{8}=2 \sqrt{2} \text {. }
$$

Therefore, the answer is $(D)$.