$20 \cdot 117$ A circle of radius $r$ is enclosed by a circle of radius $R$. The area of the larger circle is $\frac{a}{b}$ times the area of the region between the two circles $(a>b)$, then $R: r$ equals
(A) $\sqrt{a}: \sqrt{b}$.
(B) $\sqrt{a}: \sqrt{a-b}$.
(C) $\sqrt{b}: \sqrt{a-b}$.
(D) $a: \sqrt{a-b}$.
(E) $b: \sqrt{a-b}$.
(20th American High School Mathematics Examination, 1969)
[Solution] Let the radii of the two circles be $R$ and $r$. According to the given conditions, we have
$$
\pi R^{2}=\frac{a}{b}\left(\pi R^{2}-\pi r^{2}\right) \text {, or } \frac{a}{b} r^{2}=R^{2}\left(\frac{a}{b}-1\right) \text {. }
$$

Then $\frac{R^{2}}{r^{2}}=\frac{a}{a-b}$, which means $\frac{R}{r}=\frac{\sqrt{a}}{\sqrt{a-b}}$. Therefore, the answer is $(B)$.