For the question stated in the title, the answer is yes, if more is interpreted as "more than or equal to".
Proof: let $\Lambda$ be a collection of lines, and let $P$ be the extended two plane (the Riemann sphere). Let $P_1$ be a connected component of $P\setminus \Lambda$. Let $C$ be a small circle entirely contained in $P_1$. Let $\Phi$ be the conformal inversion of $P$ about $C$. Then by elementary properties of conformal inversion, $\Phi(\Lambda)$ is now a collection of circles in $P$. The number of connected components of $P\setminus \Phi(\Lambda)$ is the same as the number of connected components of $P\setminus \Lambda$ since $\Phi$ is continuous. So this shows that for any collection of lines, one can find a collection of circles that divides the plane into at least the same number of regions.
Remark: via the conformal inversion, all the circles in $\Phi(\Lambda)$ thus constructed pass through the center of the circle $C$. One can imagine that by perturbing one of the circles somewhat to reduce concurrency, one can increase the number of regions.
Another way to think about it is that lines can be approximated by really, really large circles. So starting with a configuration of lines, you can replace the lines with really really large circles. Then in the finite region "close" to where all the intersections are, the number of regions formed is already the same as that coming from lines. But when the circles "curve back", additional intersections can happen and that can only introduce "new" regions.
Lastly, yes, the number you derived is correct. See also this OEIS entry.
I have a partial solution excluding tangenciality, an upper bound for the setups.
Basically you can divide the relations between circles in three excluding types: inclusion, exclusion and intersection, i.e., a circle can cut other in two points, can be completely separated of other or it can be into some one.
If some circle is into the other both circles have the same property of exclusion respect to an exterior circle, i.e., $(A\cap B=\emptyset)\land (C\subset B)\implies C\cap A=\emptyset$.
Tangenciality can be thought as a particular case for inclusion or exclusion. Try to see the circles as open sets.
So the properties of inclusion or exclusion are "inherited" by subsets. I know an analogy on number theory: divisibility and co-primality. Some number can be co-prime of other OR can divide one to the other (in some direction) OR neither, this is analogous to inclusion/exclusion/intersection from before.
So I have some number N of circles and I want to characterize every circle with a number completely unique. Each number will be composed by some combination of factors.
I need, at least, one prime factor completely different for each number so I can have all numbers completely different one of each other if I want.
UPDATE:
I discovered that you need, for some setup of circles, more than just N factors. You cant express all possible configurations with just N factors, by example in a chain of circles. Example: for 3 circles, linked like a chain (one cut other in a line) you need at least $ab$, $bc$ and $cd$... so you need 4 factors for 3 circles.
I can represent a valid setup with a (0,1)-matrix, where every row represent a circle and every column represent existence of some same prime factor. So a setup of circles S are all the valid and not redundant squared (0,1)-matrix with dimension N, where N is the number of circles:
$$\mathbf{S_{N\times N}} = \left( \begin{array}{rrrr} 1 & x & \cdots & x \\ x & 1 & \cdots & x \\ \vdots & \vdots & \ddots & \vdots \\ x & x & \cdots & 1 \end{array} \right).$$
where $x\in \{0, 1\}$
A matrix S is valid if:
- It have the diagonal full of ones.
- No one row is identical to other (the circles are different circles).
But the big problem here is redundancy: a matrix is redundant to other if it is equivalent to any other trough any number of rows and columns swap.
And here is where lies the big problem to count these matrices: the classes of equivalence for a (0,1)-matrix under swaps of rows and columns is analogous to the graph isomorphism problem [1] [2].
So we have, IMO, two main strategies here: try to define some bound for these number of matrices (and if possible include tangenciality) or just evaluate these numbers trough experimentation with some statistical program like R.
In any case it seems that doesnt exist a closed form to evaluate the number of these setups with circles.
P.S.: alternatively you can represent setup of circles with a complete graph.
Best Answer
HINT: Suppose that $n$ circles divide the plane into $r_n$ regions. Now you add another circle, $C$, that non-trivially intersects each of the first $n$ circles. Start at any point on $C$ and travel around $C$; you will cross one of the other $n$ circles $2n$ times. If $p_0$ and $p_1$ are two consecutive crossings, the arc of $C$ from $p_0$ to $p_1$ divides one of the original $r_n$ regions into two smaller regions. You do this $2n$ times; how many new regions does that add? That gives you a recurrence $$r_{n+1}=r_n+\text{some simple function of }n\;,$$
and you can fairly easily turn it into an expression for $r_n$ that is just a summation. If you do it right, it’s a summation that you can evaluate pretty easily in closed form to get a formula for $r_n$. If you get stuck, mouse over the spoiler-protected block below for a further hint.