Circle Through Three Points

Question

Given the Cartesian coordinates of three points on a plane, find the equation of the circle through them all. The three points will not be on a straight line.

Each line of input to your program will contain the x and y coordinates of three points, in the order A(x),A(y),B(x),B(y),C(x),C(y). These coordinates will be real numbers less than 1,000,000 separated from each other by space.

The solution is to be printed as an equation of the form (x-h)^2 + (y-k)^2 = r^2. Values for h, k, r are to be printed with three digits after the decimal point. Plus and minus signs in the equations should be changed as needed to avoid multiple signs before a number.

Sample Inputs

7.0 -5.0 -1.0 1.0 0.0 -6.0
1.0 7.0 8.0 6.0 7.0 -2.0

Sample Outputs

(x - 3.000)^2 + (y + 2.000)^2 = 5.000^2
(x - 3.921)^2 + (y - 2.447)^2 = 5.409^2

Answer 1

Python, 176 189 chars

import sys,re
for s in sys.stdin:x,y,z=eval(re.sub(r'(\S+) (\S+)',r'\1+\2j,',s));w=z-x;w/=y-x;c=(x-y)*(w-abs(w)**2)/2j/w.imag-x;print'(x%+.3f)^2+(y%+.3f)^2=%.3f^2'%(c.real,c.imag,abs(c+x))

Does all its work in the complex plane. I go the math from the bottom of this page. -c is the center of the circle.

Answer 2

C# - 490

using System;class C{static void Main(){Func<string,double>p=s=>double.Parse(s);Func<double,string>t=s=>(s<0?"+ ":"- ")+Math.Abs(s).ToString("F3");foreach(var l in System.IO.File.ReadAllLines("i")){var v=l.Split();double a=p(v[0]),b=p(v[1]),c=p(v[2]),d=p(v[3]),e=p(v[4]),f=p(v[5]),m=(d-b)/(c-a),n=(f-d)/(e-c),x=(m*n*(b-f)+n*(a+c)-m*(c+e))/(2*(n-m)),y=-(x-(a+c)/2)/m+(b+d)/2,r=Math.Sqrt((x-a)*(x-a)+(y-b)*(y-b));Console.WriteLine("(x "+t(x)+")^2+(y "+t(y)+")^2 = "+r.ToString("F3")+"^2");}}}

This finds the 2 lines between AB and BC. Then it finds where the bisects of those 2 lines intersect. (Which I just noticed is what @PeterTaylor mentioned in his comment to @PeterOfTheCorn.)

Answer 3

Ruby, 192 characters

$<.map{|l|a,b,c,d,e,f=l.split.map &:to_f
n=(f-d)/(e-c)
puts"(x%+.3f)^2+(y%+.3f)^2=%.3f^2"%[x=-(n*(a+c)+(n*(b-f)-(c+e))*m=(d-b)/(c-a))/2/n-=m,y=-(x+(a+c)/2)/m-(b+d)/2,((a+x)**2+(b+y)**2)**0.5]}

Usage examples:

$ echo "7.0 -5.0 -1.0 1.0 0.0 -6.0
1.0 7.0 8.0 6.0 7.0 -2.0" | ruby circle.rb
(x-3.000)^2+(y+2.000)^2=5.000^2
(x-3.921)^2+(y-2.447)^2=5.409^2

Answer 4

Wolfram Alpha (27)

I say, use the proper tool for the job.

equation circle ([Input1],[Input2]),([Input3],[Input4]),([Input5],[Input6])

Example here.

Answer 5

Javascript (299)

The only way that I could think of solving this was algebraically solving three equations for three unknowns to find h, k, and r.

p=prompt().split(' ');a=p[0],b=p[1],c=p[2],d=p[3],e=p[4],f=p[5];h=((a*a+b*b)*(f-d)+(c*c+d*d)*(b-f)+(e*e+f*f)*(d-b))/(a*(f-d)+c*(b-f)+e*(d-b))/2;k=((a*a+b*b)*(e-c)+(c*c+d*d)*(a-e)+(e*e+f*f)*(c-a))/(b*(e-c)+d*(a-e)+f*(c-a))/2;r=Math.sqrt((a-h)*(a-h)+(b-k)*(b-k));alert("(x-"+h+")²+(y-"+k+")²="+r+"²");

Example I/O:

7.0 -5.0 -1.0 1.0 0.0 -6.0 --> (x-3)²+(y--2)²=5²

1.0 7.0 8.0 6.0 7.0 -2.0 --> (x-3.9210526315789473)²+(y-2.4473684210526314)² =5.409159155551175²

The only bug that I see is that if h or k is negative, it outputs -- instead of +.