At a high level, they are reformulating the problem so that it is easier to solve. By casting in the light below, they limit the number of possible lines to consider.
Problem A: There are N circles. The center of the i-th circle is (xi,
yi) and all circles have the radii R. And let we can draw X lines such
that any circle intersects with at least one line. What is the minimal
X?
To explain further, lets rephrase the problem A in words: The restaurants are sticklers for rules and there is a rule that says all restaurants must agree on a single maximum distance from the road - this'll be R. The circle created by the restaurant and R represents the place where a line needs to intersect to satisfy this requirement. The new problem asks the minimum number of roads to do this.
If this is not possible in under K roads, then something has to change. We can't add roads per the original problem, but we can modify R. This is where binary search comes in, but we have to solve problem A first.
Now, let's consider solving Problem A. At first, the lines can be
limited to common tangents to two circles. Because if a line
intersects with some (at least 2) circles, we can move the line such
that a moved line intersects with the same circles, and the moved line
is one of common tangents. If a line intersects less than 2 circles,
it is useless (but be careful of the case N = 1). There are at most 4
lines that is common tangents to two circles, so we consider at most 2
* N * (N-1) lines.
The important part is this, we need to find lines that intersect more than one circle. At most four lines from each pair of circle need be considered, check the source codes for implementation.
The next big step is the dynamic programming which find the minimum number of lines to cover all the circles. The 'mask' is a bitmask indicating which circles have been hit as each line is considered.
This solves the problem, but now we have to convert back. Remember R? We can now binary search to find the minimum R such that X<=K. In terms of my reformulation of Problem A, its the smallest distance all restaurant will agree to and still be serviced by a road
Hope that helps, tricky, but interesting problem.