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 forms a set $A$. The area of the figure $S$ represented by $A$ in the $a 0, b$ plane is $\qquad$
7. $\pi$
The array $(a, b)$ that makes the equation $x^{2}+2 b x+1=2 a(x+b)$ have no real roots $x$ satisfies the requirement,
that is, the condition for the equation $x^{2}+2(b-a) x+1-2 a b=0$ to have no real roots.
Thus, $\Delta=[2(b-a)]^{2}-4(1-2 a b)<0$, from which we get $a^{2}+b^{2}<1$.
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 $S$ representing it is $\pi \cdot 1^{2}=\pi$.