Consider the two functions $f(x) = x^2+2bx+1$ and $g(x) = 2a(x+b)$, where the variable $x$ and the constants $a$ and $b$ are real numbers. 
Each such pair of constants $a$ and $b$ may be considered as a point $(a,b)$ in an $ab$-plane. 
Let $S$ be the set of such points $(a,b)$ for which the graphs of $y = f(x)$ and $y = g(x)$ do not intersect (in the $xy$-plane). The area of $S$ is 
$\textbf{(A)} \ 1 \qquad  \textbf{(B)} \ \pi \qquad  \textbf{(C)} \ 4 \qquad  \textbf{(D)} \ 4 \pi \qquad  \textbf{(E)} \ \text{infinite}$
We must describe geometrically those $(a,b)$ for which the equation $x^2+2bx+1=2a(x+b)$, i.e. $x^2+2(b-a)x+(1-2ab)=0$, has no solutions (equivalent to the graphs not intersecting). By considering the discriminant of this quadratic equation, there are no solutions if and only if $\left(2(b-a)\right)^2 - 4(1)(1-2ab) < 0 \Rightarrow (b-a)^2 < 1 - 2ab \Rightarrow a^2 - 2ab + b^2 < 1 - 2ab \Rightarrow a^2 + b^2 < 1$. Thus $S$ is the unit circle (without its boundary, due to the inequality sign being $<$ rather than $\leq$, but this makes no difference to the area), whose area is $\pi (1^2) = \pi$, so the answer is $\boxed{\textbf{(B)} \ \pi}$.