TITLE: Determine if circle is covered by some set of other circles
QUESTION [20 upvotes]: Suppose you have a set of circles $\mathcal{C} = \{ C_1, \ldots, C_n \}$ each with a fixed radius $r$ but varying centre coordinates. Next, you are given a new circle $C_{n+1}$ with the same radius $r$ as the previous circles but with a new centre coordinate.
How can you determine whether the area covered by the $C_{n+1}$ is fully covered by $\mathcal{C}$?
How to do this if the circles can have varying radii?
Note: I couldn't yet work out a nice mathematical solution for this. Coming from computer science, the best I could come up with is solving this in a nasty brute force way using some sort of Monte Carlo sampling, i.e. draw a large number of random points from the area of $C_{n+1}$ and then checking for each point if it is enclosed by at least one circle in $\mathcal{C}_{\text{intersecting}}$ (subset of $\mathcal{C}$ with circles that are within $2r$ of $C_{n+1}$).

REPLY [6 votes]: The problem (for discs of arbitrary radii) is essentially equivalent to the following: given a convex polyhedral set in $\mathbb R^3$ (represented by a system of linear inequalities), determine whether it intersects the sphere $x^2+y^2+z^2=1$. To solve this, one can check if the set is compact and enumerate its vertices to see whether it lies inside the sphere, and minimize $x^2+y^2+z^2$ over the set to find out whether it is completely outside. I am not an expert in complexity but I believe that the former can be done in $O(n^2)$ time and the latter is even faster.
The reduction goes as follows. Construct a stereographic projection from the plane to the sphere in such a way that the main disc $C_{n+1}$ is mapped to the upper hemisphere. Every other disc is mapped to spherical cap which can be represented as the set of solution of a linear inequality (whose coefficients are easy to compute) intersected with the sphere. Add the inequality $z\le 0$ for the lower hemisphere. Now we need to find out whether the union of corresponding half-spaces cover the sphere. Or, equivalently, whether the intersection of their complement half-spaces have an empty intersection with the sphere. Q.E.D.
The fact that the spherical caps arise from discs in the plane (i.e., not from complements of discs) means that the complement half-spaces except the last one contain the south pole $(0,0,-1)$. However this does not help much. If the original radii are equal, then the complement half-spaces also contain the center $(0,0,0)$. This rules out the case when the intersection is outside the sphere and thus simplifies the problem a little bit.