What is maximum number of circles in $\mathbb R^2$ that intersect a common circle, but are otherwise disjoint? Given a circle of a fixed radius in the plane, how many circles can be drawn intersecting with that circle, that are disjoint from each other?
(If this type of question has a name or has been studied in the past, feel free to let me know! It seems vaguely related to sphere-packing stuff, but not exactly)
More precisely, let $S^1 \subset \mathbb R^2$ be the unit circle entered at the origin. What is the maximum number of circles $S_i$ of radius $1$ such that $S_i \cap S^1 \neq \emptyset$ for all $i$, and $S_i \cap S_j = \emptyset$ for all $i\neq j$?
I spent some time drawing examples and started with squares, since they are a lot easier to draw. I suspect the answer is 5 for squares, but I'm not totally sure, and would like to know how to go about proving this rigorously.
Remark (edit): I also know about doubling dimension, which may have some role to play in the answer (emphasis on "may"!) but of course the definition is quite dissimilar from my problem is that it uses balls of radius $r/2$, and does not care whether the covering balls intersect, though maybe the doubling constant gives a weak upper bound or something. (Brian below thinks it is $7$.)
Remark 2: Since I forgot about rotations, my answer of 5 for squares was not correct. Coincidentally, it ended up being correct for circles (maybe indirectly because I was drawing squares with the same orientation). At any rate, if anyone can find the answer for squares as well, feel free, or maybe I should post as a separate question.
 A: Let $O$ be the centre of your original circle.
Let $O_1$ and $O_2$ be the centres of any two such circles. As the circles don't intersect each other, it follows that $|O_1O_2|>2$. However, they intersect the original circle, so $|OO_1|,|OO_2|\le 2$. This implies that, in the triangle $\triangle OO_1O_2$ the side $O_1O_2$ is the longest, so the opposite angle $\angle O_1OO_2$ must be the largest, and therefore larger than $60^\circ$.
Now, join $O$ with all the centres $O_i$ via rays. Any two of those rays make up an angle $>60^\circ$, and so those rays divide the full angle of $360^\circ$ into angles $>60^\circ$, which can only happen if there are at most five of them.
It is also easy to show that you can draw five circles to satisfy your conditions. (Imagine a regular pentagon inscribed in your circle, and five other circles touching at the corners of the pentagon.)
So, five is your answer.
A: The unit circles centered at two points $a$ and $b$ intersect iff $d(a,b)\leq 2$.  So, you are equivalently asking how many points you can have inside a disk of radius $2$ such that the distance between any two of them is greater than $2$.  Or, scaling down, you are asking how many points you can have in a unit disk such that the distance between any two of them is greater than $1$.
Let us consider the unit disk $D\subset\mathbb{C}$ centered at $0$.  I claim that if $a,b\in D$ have arguments that differ by at most $\pi/3$, then $|a-b|\leq 1$.  We may assume $0\leq a\leq 1$.  First, if $a=1$ and $|b|=1$ with $b=e^{i\theta}$, then $|a-b|^2=2-2\cos(\theta)\leq 1$ if $\theta\in[-\pi/3,\pi/3]$.  Since $|a-0|\leq 1$ as well, it follows by convexity of disks that the unit disk around $a$ contains all points of $D$ whose argument is in $[-\pi/3,\pi/3]$.  Finally, if $0\leq a\leq 1$ and $b\in D$ has argument in $[-\pi/3,\pi/3]$, then $0$ and $1$ are both within distance $1$ of $b$ and thus so is $a$, again by convexity.
So, given a subset of $D$ all of whose elements are a distance greater than $1$ apart, their arguments must all be greater than $\pi/3$ apart.  It follows that there can be at most $5$ such points.  On the other hand, it is easy to find $5$ points that work (for instance, $e^{2\pi i k/5}$ for $k=0,1,2,3,4$).
