$26 \cdot 12$ The function $f(x)=x^{2}+2 b x+1$ and $g(x)=2 a(x+b)$, where the variable $x$ and constants $a, b$ are real numbers. Each pair of constants $a, b$ can be considered as a point $(a, b)$ in the $ab$-plane. Let $S$ be the set of these points $(a, b)$ for which the curves $y=$ $f(x)$ and $y=g(x)$ do not intersect in the $xy$-plane. The area of $S$ is
(A) 1 .
(B) $\pi$.
(C) 4 .
(D) $4 \pi$.
(E) infinite.
(34th American High School Mathematics Examination, 1983)
[Solution] From the problem, the points $(a, b)$ in the set $S$ satisfy the equation for $x$
$$
x^{2}+2 b x+1=2 a(x+b)
$$

having no real roots, which means the equation
$$
x^{2}+2(b-a) x+(1-2 a b)=0
$$

has no real roots.
Thus,
$$
\begin{aligned}
\Delta= & {[2(b-a)]^{2}-4(1-2 a b)<0, } \\
& 4\left(a^{2}-2 a b+b^{2}\right)-4+8 a b<0, \\
& 4 a^{2}+4 b^{2}<4, \text { i.e., } a^{2}+b^{2}<1 .
\end{aligned}
$$

Therefore, $S$ is an unbounded unit circle, with an area of $\pi$.
Hence, the answer is $(B)$.