The answer below involves pretending that the earth is a perfect sphere, which should give a more accurate answer than treating the earth as a flat plane.
To figure out the radius of a set of lat/lon points, you must first ensure that your set of points is "hemispherical", ie. all the points can fit into some arbitrary half of your perfect sphere.
Check out Section 3 in the paper "Optimal algorithms for some proximity problems on the Gaussian sphere with applications" by Gupta and Saluja. I don't have a specific link, but I believe that you can find a copy online for free. This paper is not sufficient to implement a solution. You'll also need Appendix 1 in "Approximating Centroids for the Maximum Intersection of Spherical Polygons" by Ha and Yoo.
I wouldn't use Megiddo's algorithm for doing the linear programming part of the hemisphericity testing. Instead, use Seidel's algorithm for solving Linear Programming problems, described in "Small-Dimensional Linear Programming and Convex Hulls Made Easy" by Raimund Seidel. Also see "Seidel’s Randomized Linear Programming Algorithm" by Kurt Mehlhorn and Section 9.4 from "Real-Time Collision Detection" by Christer Ericson.
Once you have determined that your points are hemispherical, move on to Section 4 of the paper by Gupta and Saluja. This part shows how to actually get the "smallest enclosing circle" for the points.
To do the required quadratic programming, see the paper "A Randomized Algorithm for Solving Quadratic Programs" by N.D. Botkin. This tutorial is helpful, but the paper uses (1/2)x^T G x - g^T x and the web tutorial uses (1/2)x^T H x + c^T x. One adds the terms and the other subtracts, leading to sign-related problems. Also see this example 2D QP problem. A hint: if you're using C++, the Eigen library is very good.
This method is a little more complicated than some of the 2D methods above, but it should give you more accurate results than just ignoring the curvature of the earth completely. This method also has O(n) time complexity, which is likely asymptotically optimal.
Note: The method described above may not handle duplicate data well, so you may want to check for duplicate lat/lon points before finding the smallest enclosing circle.