If K is not large enough and you need to find a point with integer coordinates, you should do, as another answer suggested - Calculate minimum distances for all points on the grid, using BFS, strarting from all given points at once.

Faster solution, for large K, and probably the only one which can find a point with float coordinates, is as following. It has complexity of O(n log n log k)

Search for resulting maximum distance using dihotomy. You have to check if there is any point inside the square [0, k] X [0, k] which is at least given distance away from all points in given set. Suppose, you can check that fast enough for any distance. It is obvious, that if there is such point for some distance R, there always will be some point for all smaller distances r < R. For example, the same point would go. Thus you can search for maximum distance using binary search procedure.

Now, how to fast check for existence (and also find) a point which is at least r units away from all given points. You should draw "Manhattan spheres of radius r" around all given points. These are set of points at most r units away from given point. They are tilted by 45 degrees squares with diagonal equal to 2r. Now turn the picture by 45 degrees, and all squares will be parallel to the axis. Now you can check for existence of any point outside such squares using sweeping line algorithm. You have to sort all vertical edges of squares, and then process them one by one from left to right. Left borders will add segment mark to sweeping line, Left borders will erase it. And you have to check if there is any non marked point on the line. You can implement it using segment tree. Then, you have to check if there is any non marked point on the line inside the initial square [0,k]X[0,k].

So, again, overall solution will be binary search for r. Inside of it you will have to check if there is any point at least r units away from all given points. Do that by constructing "manhattans spheres of radius r" and then scanning them with a diagonal line from left-top corner to right-bottom. While moving line you should store number of opened spheres at each point at the line in the segment tree. between opening and closing of any spheres, line does not change, and if there is any free point there, it means, that you found it for distance r.

Binary search contributes log k to complexity. Each checking procedure is n log n for sorting squares borders, and n log k (n log n?) for processing them all.

Voronoi diagram would be another fast solution and could also find non integer answer. But it is much much harder to implement even for Manhattan measure.