7. 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 $\qquad$.
Let the pair of numbers $(a, b)$ satisfy the requirement, such that the equation $x^{2}+2 b x+1$ $=2 a(x+b)$ has no real roots.
$$
\begin{array}{l}
\therefore \Delta=[2(b-a)]^{2}-4(1-2 a b)<0, \text { which gives } \\
a^{2}+b^{2}<1 .
\end{array}
$$

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 circle). 
Therefore, the area of the figure $S$ is $\pi \times 1^{2}=\pi$.