<p>You have two distinct problems: you need to turn the geometric problem into a combinatoric problem, and then you need to solve the combinatoric problem. For the latter, you are looking at a <a href="https://en.wikipedia.org/wiki/Set_cover_problem" rel="nofollow noreferrer">minimum set cover problem</a>, and there should be plenty of literature on that. Personally I like Knuth's <a href="https://arxiv.org/abs/cs/0011047" rel="nofollow noreferrer">Dancing Links</a> approach to enumerate all solutions of a set cover, but I guess for a single minimal solution you can do better. A CPLEX formulation (to match your tag) would use a binary variable for each row, and a ≥1 constraint for each column.</p>

<p>So now about turning geometry into combinatorics. All the lines of all your circles divide the plane into a bunch of areas. The areas are delimited by lines. Of particular relevance are the points where two or more circles meet. The exact shape of the line between these points is less relevant, and you might imagine pulling those arcs straight to come up with a more classical planar graph representation. So compute all the pair-wise intersections between all your circles. Order all intersections of a single circle <a href="https://stackoverflow.com/q/16542042/1468366">by angle</a> and connect them with graph edges in that order. Do so for all circles. Then you can do a kind of bucket fill to determine for each circle which graph faces are within and which are outside.</p>

<p>Now you have your matrix for the set cover: every graph face which is inside the big circle is a column you need to cover. Every circle is a row and covers some of these faces, and you know which.</p>
How to select the minimal set of circles that covers another circle?
<p>I'm looking for some solutions that, given a set <code>S</code> of circles with 2D-center points and radii, returns a minimal sub-set <code>M</code> in <code>S</code> that covers entirely a specific circle with 2d-center point and radius. This last circle is not in <code>S</code>.</p>

<p>I've chosen circles, but it doesn't matter if we change them to squares, hexagons, etc.</p>