20.108 As shown in the figure, two circles are externally tangent, and the lines $P A B$ and $P A' B'$ are common tangents to the two circles, intersecting the smaller circle at $A$ and $A'$, and the larger circle at $B$ and $B'$. If $P A = A B = 4$, then the area of the smaller circle is
(A) $1.44 \pi$.
(B) $2 \pi$.
(C) $2.56 \pi$.
(D) $\sqrt{8} \pi$.
(E) $4 \pi$.
(42nd American High School Mathematics Examination, 1991)
[Solution] Let the center of the large circle be $O$, with radius $R$, and the center of the small circle be $O^{\prime}$, with radius $r$. Connect $A O^{\prime}, B O$, and $P O^{\prime} O$. Given that $P A=A B$, then $O^{\prime}$ is the midpoint of $P O$, and $R=$
$2 r$.
Also, $P B=P A+A B=8, O B=2 r$,
$P O=2 O O^{\{\prime}=2(R+r)=2(2 r+r)=6 r$,
then by the Pythagorean theorem $P O^{2}=O B^{2}+P B^{2}$,
i.e., $64+4 r^{2}=36 r^{2}$, we get $r^{2}=2$.
Thus, the area of the small circle is $2 \pi$.
Therefore, the answer is $(B)$.