The problem:

Given a non-empty set of points in the cartesian plane, find the smallest circle that encloses them all (Wikipedia link).

This problem is trivial if the number of points is three or less (if there's one point, the circle has a radius of zero; if there are two points, the line segment that joins the points is the diameter of the circle; if there are three (non co-linear) points, it's possible to get the equation of a circle that touches them all if they form a non-obtude triangle, or a circle that touches only two points and encloses the third if the triangle is obtuse). So, for the sake of this challenge, the number of points should be greater than three.

The challenge:

• 
  Input: A list of 4 or more non colinear points. The points should have X and Y coordinates; coordinates can be floats. To ease the challenge, no two points should share the same X coordinate.
  
  For example: [(0,0), (2,1), (5,3), (-1,-1)]
  


• 
  Output: A tuple of values, (h,k,r), such that $(x-h)^2 + (y-k)^2 = r^2$ is the equation of the smallest circle that encloses all points.

Rules:

• You can choose whatever input method suits your program.


• Output should be printed to STDOUT or returned by a function.


• "Normal", general-purpose, languages are preferred, but any esolang is acceptable.


• You can assume that the points are not colinear.


• Output can be rounded to 2 decimal position (if that eases your work).


• This is code-golf, so the smallest program in bytes wins. The winner will be selected one week after the challenge is posted.
  
  
  
  
  • Please include the language you used and the length in bytes as header in the first line of your answer: # Language: n bytes
    
  
  
  
  

Test cases:

• 1:
  
  
  
  
  • Input: [(-8,0), (3,1), (-6.2,-8), (3,9.5)]
    
  
  
  • Output: [-1.6, 0.75, 9.89]
    
  
  
  
  


• 2:
  
  
  
  
  • Input: [(7.1,-6.9), (-7,-9), (5,10), (-9.5,-8)]
    
  
  
  • Output: [1.65, 0.52, 11.58]
    
  
  
  
  


• 3:
  
  
  
  
  • Input: [(0,0), (1,2), (3,-4), (4,-5), (10,-10)]
    
  
  
  • Output: [5.5, -4.7, 7.5]
    
  
  
  
  


• 4:
  
  
  
  
  • Input: [(6,6), (-6,7), (-7,-6), (6,-8)]
    
  
  
  • Output: [0, -0.5, 9.60]
    
  
  
  
  

Happy golfing!!!

Related challenge:

• Area of a 2D convex hull

Wolfram Language (Mathematica), 28 bytes

BoundingRegion[#,"MinDisk"]&

Try it online!

Built-ins are handy here. Output is a disk object with the centre and radius. Like others, I’ve found the 2nd and 3rd test cases to be different to the question.

R, 67 bytes

function(x,a=nlm(function(y)max(Mod(x-y%*%c(1,1i))),0:1))c(a$e,a$m)

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Takes input as a vector of complex coordinates. Mod is the distance (modulus) in the complex plane and nlm is an optimization function: it finds the position of the center (a$e) which minimizes the maximum distance to the input points, and gives the corresponding distance (a$m), i.e. the radius. Accurate to 3-6 digits; the TIO footer rounds the output to 2 digits.

nlm takes a numeric vector as input: the y%*%c(1,1i) business converts it to a complex.

JavaScript (ES6), 298 bytes

Returns an array [x, y, r].

p=>p.map(m=h=>p.map(i=>[0,...p].map(j=>!p.every(([x,y])=>H(x-X,y-Y)<=r,([a,A]=h,[b,B]=i),X=j?x=([c,C]=j,P=a*a+A*A,Q=b*b+B*B,R=c*c+C*C,d=c*(l=A-B)+a*(B-=C)+b*(C-=A),P*B+Q*C+R*l)/d/2:(a/=2)+(x=b/2),Y=j?y=(P*(c-b)+Q*(a-c)+R*(b-a))/d/2:(A/=2)+(y=B/2),H=Math.hypot,r=H(x-a,y-A))|r>m||(o=[X,Y,m=r]))))&&o

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Jelly, 100 bytes

ZÆm,Hñ/$

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1ịṭƊZIṚṙ€1N1¦,@ṭ@²§$µḢZḢ×Ø1œị$SḤ÷@P§

ZṀ+ṂHƲ€_@Ç+ḷʋ⁸,_²§½ʋḢ¥¥

œc3Ç€;ŒcÑ€$2ŀ_ƭƒⱮṀ¥Ðḟ⁸ṚÞḢ

Try it online!

In contrast to my Wolfram language answer, Jelly needs quite a lot of code to achieve this (unless I’m missing something!). This full program takes the list of points as its argument and returns the centre and radius of the smallest enclosing circle. It generates circumcircles for all sets of three points, and diameters for all sets of two points, checks which include all of the points and then picks the one with the smallest radius.

Code for the make_circumcircle bit was inspired by code at this site, in turn inspired by Wikipedia.