This question already has an answer here:
Yes, I’ve seen this one before. Assuming exactly one penny is allowed to be placed per turn:
Go first and place a penny in the dead center of the table. From then on, any move your opponent makes, place a penny in the mirror opposite location (i.e. rotated 180 degrees). It stands to reason that if your opponent’s move was valid, yours will be too. Hence, you will always have an available move if your opponent does. Since the table is only finitely large, there can only be finitely many turns, hence you will eventually win.
A more complete proof:
Suppose the table is described using polar coordinates with the center of the table as the origin ($r=0$).
My first move is to place at $r=0$. When my opponent makes a legal move at $(r,\theta)$ I attempt to place a coin at $(r,\theta+180^\circ)$.
Claim: I am always allowed to do so and such a move will always be valid.
Proof: Suppose otherwise. Then that implies that either the target location is not on the table (in which case my opponent’s previous move will also have not been on the table and therefore was also invalid), or that target location would have a coin overlap with another previously placed coin. As it could not have been the coin that my opponent has just placed on his last turn (as it is $180^\circ$ away), that implies that those coins must have been placed previously. However… since my moves are always playing $180^\circ$ away from my opponent, that implies that there should be the same situation on the other side of the table and that my opponents coin also is overlapping the corresponding mirrored coins and therefore my opponents move was invalid. Either way, we reach a contradiction implying that if my opponents move was valid that my move is also guaranteed to be valid too.
JMoravitz has a great post. I’m happy to have arrived at the same answer, and would like to share a thought that aided my process. Consider an extreme case to find out whether to go first or second:
The problem specifies pennies and a circular table. If there is a winning strategy, then without any loss of generality, we ought to win regardless of the radius of the table. Given a table whose radius is smaller than the penny’s, and assuming that at least one penny would fit on the table, then you clearly want to go first. You would immediately win.
What if the table was… well… normal? How to ensure your win? Don’t run out of options, of course. This is the same idea as J’s in the other answer.
Given the centrosymmetric nature of the arrangement, after centering your original penny, player two can now be mimicked until the game’s end. For any penny the opponent places at $(x,y)$, place your next one at $(-x,-y)$. I’d expect most opponents to resign after a turn or two, or suffer a slow, sad defeat.
Zugzwang after move 1 really.
I’m wondering if it would be possible to prove that the following solution is a winning or losing strategy:
Go first and place a penny near the edge of the table in such a way that there isn’t room for another penny on the outside of it.
Update: OP changed rules at revision 3.
My attempt as spoiler below.
I move first and cover the whole table with pennies.