1. Given the functions $f(x)=x^{2}+2 b x+1$ and $g(x)=2 a(x+b)$, where $x$, $a$, and $b$ are real numbers, such that the graphs of $y=f(x)$ and $y=g(x)$ do not intersect in the $x O y$ plane, the set of real number pairs $(a, b)$ that satisfy this condition forms a point set $A$. Then, the area of the figure $T$ represented by $A$ in the $a O b$ plane is $\qquad$
(2002, Anhui Province High School Mathematics Competition)
(Due to the pair of numbers $(a, b)$ making the equation $x^{2}+2 b x+1=$ $2 a(x+b)$ have no real roots, thus, $\Delta<0$, which means $a^{2}+b^{2}<1$ represents the area inside the circle. The area of figure $T$ is $\pi$.)