求解方程从3个点找到圆心

Question

I'm looking for a high precision solution to find the center point of a circle from 3 data points on a canvas (x,y). I found this example in the attached screenshot above, now I'm using the Math.NET package to solve the equation and I'm comparing the results against this online tool: https://pl.netcalc.com/8116/ .

However, when I calculate the radius its completely off and often a negative number???

using MathNet.Numerics.LinearAlgebra.Double.Solvers;
using MathNet.Numerics.LinearAlgebra.Double;
using System;

namespace ConsoleAppTestBed
{
  class Program
  {
    static void Main(string[] args)
    {
        var dataPoints = new double[,]
        {

            { 5, 80 },
            { 20, 100 },
            { 40, 140 }
        };


        var fitter = new CircleFitter();
        var result = fitter.Fit(dataPoints);

        var x = -result[0];
        var y = -result[1];
        var c = result[2];

        Console.WriteLine("Center Point:");
        Console.WriteLine(x);
        Console.WriteLine(y);
        Console.WriteLine(c);

        //// (x^2 + y^2 - c^2)
        var radius = Math.Pow(x, 2) + Math.Pow(y, 2) - Math.Pow(c, 2);
        //// sqrt((x^2 + y^2 - c^2))
        radius =  Math.Sqrt(radius);
        Console.WriteLine("Radius:");
        Console.WriteLine(radius);
        Console.ReadLine();
    }

    public class CircleFitter
    {
        public double[] Fit(double[,] v)
        {
            var xy1 = new double[] { v[0,0], v[0,1] };
            var xy2= new double[] { v[1, 0], v[1, 1] };
            var xy3 = new double[] { v[2, 0], v[2, 1] };

            // Create Left Side Matrix of Equation
            var a = CreateLeftSide_(xy1);
            var b = CreateLeftSide_(xy2);
            var c = CreateLeftSide_(xy3);
            var matrixA = DenseMatrix.OfArray(new[,] 
            {
                { a[0], a[1], a[2] },
                { b[0], b[1], b[2] },
                { c[0], c[1], c[2] }
            });

            // Create Right Side Vector of Equation
            var d = CreateRightSide_(xy1);
            var e = CreateRightSide_(xy2);
            var f = CreateRightSide_(xy3);
            double[] vector = { d, e, f };
            var vectorB =  Vector<double>.Build.Dense(vector);

            // Solve Equation
            var r = matrixA.Solve(vectorB);
            var result = r.ToArray();

            return result;
        }

        //2x, 2y, 1
        public double[] CreateLeftSide_(double[] d) 
        {
            return new double[] { (2 * d[0]), (2 * d[1]) , 1};
        
        }

        // -(x^2 + y^2)
        public double CreateRightSide_(double[] d) 
        { 
            return -(Math.Pow(d[0], 2) + Math.Pow(d[1], 2));

        }
    }
  }

}

Any ideas?

Thanks in advance.

Answer 1

The solution to your problem is here: The NumberDecimalDigits property

Code:

using System;
using System.Globalization;
namespace ConsoleApp1
{
class Program
{
    static void Main()
    {
        double x1 = 1, y1 = 1;
        double x2 = 2, y2 = 4;
        double x3 = 5, y3 = -3;
        findCircle(x1, y1, x2, y2, x3, y3);
        Console.ReadKey();
    }
    static void findCircle(double x1, double y1,
                           double x2, double y2,
                           double x3, double y3)
    {
        NumberFormatInfo setPrecision = new NumberFormatInfo();
        setPrecision.NumberDecimalDigits = 3; // 3 digits after the double point

        double x12 = x1 - x2;
        double x13 = x1 - x3;

        double y12 = y1 - y2;
        double y13 = y1 - y3;

        double y31 = y3 - y1;
        double y21 = y2 - y1;

        double x31 = x3 - x1;
        double x21 = x2 - x1;

        double sx13 = (double)(Math.Pow(x1, 2) -
                        Math.Pow(x3, 2));

        double sy13 = (double)(Math.Pow(y1, 2) -
                        Math.Pow(y3, 2));

        double sx21 = (double)(Math.Pow(x2, 2) -
                        Math.Pow(x1, 2));

        double sy21 = (double)(Math.Pow(y2, 2) -
                        Math.Pow(y1, 2));

        double f = ((sx13) * (x12)
                + (sy13) * (x12)
                + (sx21) * (x13)
                + (sy21) * (x13))
                / (2 * ((y31) * (x12) - (y21) * (x13)));
        double g = ((sx13) * (y12)
                + (sy13) * (y12)
                + (sx21) * (y13)
                + (sy21) * (y13))
                / (2 * ((x31) * (y12) - (x21) * (y13)));

        double c = -(double)Math.Pow(x1, 2) - (double)Math.Pow(y1, 2) -
                                    2 * g * x1 - 2 * f * y1;
        double h = -g;
        double k = -f;
        double sqr_of_r = h * h + k * k - c;

        // r is the radius
        double r = Math.Round(Math.Sqrt(sqr_of_r), 5);

        Console.WriteLine("Center of a circle: x = " + h.ToString("N", setPrecision) + 
        ", y = " + k.ToString("N", setPrecision));
        Console.WriteLine("Radius: " + r.ToString("N", setPrecision));
        }
        }
        }

Answer 2

I have just converted William Li's answer to Swift 5 for those who like to cmd+C and cmd+V like me:)

 func calculateCircle(){
    
    
    let x1:Float = 0
    let y1:Float = 0
    
    let x2:Float = 0.5
    let y2:Float = 0.5
    
    let x3:Float = 1
    let y3:Float = 0
    
    let x12 = x1 - x2
    let x13 = x1 - x3
    
    let y12 = y1 - y2
    let y13 = y1 - y3
    
    let y31 = y3 - y1
    let y21 = y2 - y1
    
    let x31 = x3 - x1
    let x21 = x2 - x1
    
    
    let sx13 = pow(x1, 2) - pow(x3, 2)

    let sy13 = pow(y1, 2) - pow(y3, 2)

    let sx21 = pow(x2, 2) - pow(x1, 2)

    let sy21 = pow(y2, 2) - pow(y1, 2)

    
    let f = ((sx13) * (x12)
                    + (sy13) * (x12)
                    + (sx21) * (x13)
                    + (sy21) * (x13))
                    / (2 * ((y31) * (x12) - (y21) * (x13)))
    
    
    let g = ((sx13) * (y12)
                    + (sy13) * (y12)
                    + (sx21) * (y13)
                    + (sy21) * (y13))
                    / (2 * ((x31) * (y12) - (x21) * (y13)))
    
    
    let c = -pow(x1, 2) - pow(y1, 2) - 2 * g * x1 - 2 * f * y1
                            
    let h = -g
    let k = -f
    let r = sqrt(h * h + k * k - c)
                            
 
    
    print("center x  = \(h)")
    print("center y = \(k)")
    print("r = \(r)")
    
    
                            
}

Answer 3

Updated Answer

The equation for radius is incorrect; it should be (not c squared):

which is why you get incorrect values for the radius.

The original answer was incorrect, but it is still 'interesting'.

(Incorrect) Original Answer

The cause of the problematic calculation is not solely due to the precision of the numbers, but more because problem is ill-conditioned. If you look at the three points and where they are located on the circle, you'll see that they are bunched on a small segment of the circumference. When the points are so close to each other, it is a tough ask to find the circle's centre and radius accurately.

So the intermediate calculations, which will have small rounding errors, result in hugely exaggerated errors.

You can see the ill-conditioned nature of the problem by adding the ConditionNumber() method.

    // Solve Equation
    Console.WriteLine(matrixA.ConditionNumber()); // <<< Returns 5800 -> Big!
    var r = matrixA.Solve(vectorB);               // Existing code

A large result indicates an ill-conditioned problem. In this case a value of 5800 is returned, which is large. You might get better results using Gaussian Elimination with partial pivoting , but it still does not address the fact that the basic problem is ill-conditioned, which is why you get wildly incorrect answers.