Let $[t]$ denote the greatest integer $\leq t$ where $t \geq 0$ and $S = \{(x,y): (x-T)^2 + y^2 \leq T^2 \text{ where } T = t - [t]\}$. Then we have 
$\textbf{(A)}\ \text{the point } (0,0) \text{ does not belong to } S \text{ for any } t \qquad$
$\textbf{(B)}\ 0 \leq \text{Area } S \leq \pi \text{ for all } t \qquad$
$\textbf{(C)}\ S \text{ is contained in the first quadrant for all } t \geq 5 \qquad$
$\textbf{(D)}\ \text{the center of } S \text{ for any } t \text{ is on the line } y=x \qquad$
$\textbf{(E)}\ \text{none of the other statements is true}$
The region $S$ is a circle radius $T$ and center $(T,0)$.  Since $T = t-[t]$, $0 \le T < 1$.  That means the area of the circle is less than $\pi$, and since the region can also be just a dot (achieved when $t$ is integer), the answer is $\boxed{\textbf{(B)}}$.