The problem is much simpler than the general Kneser-Poulsen case. Here is an elementary proof of the fact that $B/A\le r^2$ for all $r\ge 1$. To simplify technicalities, I assume that the family of circles is finite (the general case follows as a limit).
Let $p_1,\dots,p_N$ be the centers of our circles. Let $V$ denote the union of the unit circles and $U$ the union of the circles of radius $r$. Divide $U$ into Voronoi regions $U_i$ of the points $p_i$. Namely, $U_i$ is the set of points $x\in U$ such that $p_i$ is nearest to $x$ among $p_1,\dots,p_N$. (For convenience, remove the points having more than one nearest center; this is a zero measure set.) Each $U_i$ is star-shaped w.r.t. the respective point $p_i$. Indeed, if $x\in U_i$, then $|p_ix|\le r$, hence the segment $[p_ix]$ is contained in $U$. And obviously $p_i$ is the nearest center for every point of this segment.
Now apply the $(1/r)$-homothety centered at $p_i$ to each set $U_i$. The resulting sets $U_i'$ are disjoint and contained in $V$, hence
$$
S(V) \ge\sum S(U_i') = \frac1{r^2} S(U_i) = \frac1{r^2} S(U)
$$
where $S$ denotes the area. Thus $S(U) \le r^2 S(V)$, q.e.d.
The same argument works in $\mathbb R^n$ (with $r^n$ in place of $r^2$), in Alexandrov spaces of nonnegative curvature, and in Riemannian manifolds of nonnegative Ricci curvature (just combine it with the proof of Bishop-Gromov inequality).
