Example 11 (2002 Anhui Province Competition Question) Given the functions $f(x)=x^{2}+2 b x+1$ and $g(x)=2 a(x+b)$, where $x$, $a$, and $b$ are all real numbers, the set of real number pairs $(a, b)$ that make the graphs of $y=f(x)$ and $y=g(x)$ not intersect in the $x O y$ plane is denoted as $A$. Then, the area of the figure $S$ represented by $A$ in the $a O b$ plane is $\qquad$.

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Solve: Fill in $\pi$. Reason: Let the pair of numbers $(a, b)$ satisfy the requirement, i.e., the equation $x^{2}+2 b x+1=2 a(x+b)$ has no real roots.
Thus, $\Delta=[2(b-a)]^{2}-4(1-2 a b)<0$, which gives
$$
a^{2}+b^{2}<1 \text {. }
$$

In the $a O b$ plane, the set of points $(a, b)$ that satisfy this condition forms the interior region of the unit circle (excluding the boundary). Therefore, the area of the figure $S$ is $\pi \times 1^{2}=\pi$.