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Given $n$ points in 2D plane .Find the minimum radius of the circle which contains at least $k$ points ($k\leq n$)? The points can be in or on the circumference of the circle.

I tried the recursion approach; i.e, if $C(i)$ is the circle with $i$ points on or inside,but I am not able to find the relation between $C(i)$ and $C(i−1)$. Also I tried it by selecting 3 points at a time and then finding the no. of points lying within it , but it gives a complexity of $O(n^4)$. So I need an optimised solution.

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First attempt

Let the coordinates of the points be denoted like this: $(x_1,y_1),...,(x_n,y_n)$.

First calculate the center of "mass", $C$:

$$\left(\frac1n\sum_1^n x_i,\frac1n\sum_1^n y_i\right).$$

Then draw a circle of radius $r$ centered at $C$. Count the points falling inside the actual circle and increase or decrease the radius so the circle contains at least $k$ points.

Or, more like an algorithm: calculate the distance of the points from $C$. Order these distances. Then take the $k^{th}$ distance. Draw a circle centered at $C$ with that radius.

EDIT

Second attempt My solution above will not provide the minimum radius as shown below:

enter image description here

Another try:

Take the first point as a center and increase the radius until at least $k$ points will fall in the circle. Then take the next point and do the same. Finaly select the minimum of the radii you got.

I am afraid that this is still not the optimal algorithm.

Third attempt

The algorithm giving the optimal radius is very computing intensive: Take all the groups of points of $k$ elements. There are $\binom nk$ such groups. For any such group use my first algorithm. Then select the minimum radius.

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    'Take the first point as a center and increase the radius until at least k''. Is it necessary that the required circle(with min radius) should have one of the given points only, as the center? – 2017-01-09
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    No, as you see my third algorithm. The optimal solution does not necessarily use circles centered at one of the points given. I am still thinking how to optimize the algorithm... – 2017-01-09
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    n has a limit of 500 , if that helps , Combinations would render it slow ... – 2017-01-09
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    I know. This is why I am still thinking. – 2017-01-09
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    It occurs to me that that there is no better solution than the one you described in the OP. Take all the triplets and count the points falling in the actual circumcircles... Sorry, I cannot go any further. – 2017-01-09