Minimum Enclosing Circle

Problem

Return the smallest circle enclosing all points as [centre_x, centre_y, radius], or an empty list for no points.

The optimal circle is determined by either one point, a diameter through two points, or the circumcircle of three points, so test every one, two and three point candidate and keep the smallest that contains all points. Return floats.

Hint

Collinear triples have a zero denominator and can be skipped safely.

Test cases

case 1

in: [[[0,0],[2,0],[0,2],[2,2]]]

expected: [1,1,1.4142135623730951]

case 2

in: [[[0,0],[1,0]]]

expected: [0.5,0,0.5]

case 3

in: [[[5,5]]]

expected: [5,5,0]

case 4

in: [[[0,0],[3,0],[0,4]]]

expected: [1.5,2,2.5]

Starter codePython

def min_enclosing_circle(points):
    # Your code here
    pass

4 test cases · run your code against them in the editor