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Solving equation to find center point of circle from 3 points

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://planetcalc.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.

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));
}
}
}

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)")
}

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.

Not all code paths return a value, I don't know what to do

private static string GetMinX(int a, int b, int c)
{
double x1 = 0;
double x2 = 0;
double x3 = 0;
if (b * b - 4 * a * c > 0)
{
x1 = (-b - Math.Sqrt(b * b - 4 * a * c)) / 2 * a;
x2 = (-b + Math.Sqrt(b * b - 4 * a * c)) / 2 * a;
if (x2 > x1)
return x1.ToString();
else
return x2.ToString();
}
if (b * b - (4 * a * c) == 0)
{
x3 = -b / 2 * a;
return x3.ToString();
}
if (b * b - 4 * a * c < 0)
return "IMP";
}
So every condition returns value (d > 0, == 0, < 0). Why does it write not all code paths return a value?

Because the last condition if (b * b - 4 * a * c < 0) has no else case and there is no return statement following it.
This condition might always true because of your logic, but C# doesn't know that. C# just assumes that any condition can be true or false (except when the condition is a constant expression).
You have two cases:
Either this condition is supposed to be always true: then just drop the condition and always return "IMP". This is the case here.
This condition might be false sometimes: then add another return statement at the end.
private static string GetMinX(int a, int b, int c)
{
int discriminant = b * b - 4 * a * c;
if (discriminant > 0) {
double root = Math.Sqrt(discriminant);
double x1 = (-b - root) / (2 * a); // Put brackets around the divisor!
double x2 = (-b + root) / (2 * a); // Put brackets around the divisor!
if (x1 < x2)
return x1.ToString();
else
return x2.ToString();
}
if (discriminant == 0) {
double x = -b / (2 * a); // Put brackets around the divisor!
return x.ToString();
}
return "IMP";
}
It is also good to do the same calculation only once and to store the result in a variable.
There is a subtle error in ... / 2 * a. It should be ... / (2 * a), otherwise you divide by 2 and then multiply by a. But you want to divide by 2 * a.
There is no point in assigning 0 to x1, x2 and x3. These values will never be used. Better declare the variables where they are used (in the local code block) and initialize them directly with the right value. Their scope will be limited to this code block. This means that for instance x1 does not exist after the first if-statement and will therefore never have an undetermined value.
"if x1 is less than x2" sounds more natural than "if x2 is greater than x1", since you want to know which one the smallest is. But both variants will work.

I advise you to use if/else if/else statement, it should be more efficient. Also you can use Math.Min and discard x3.
private static string GetMinX(int a, int b, int c)
{
double x1, x2, m;
m = b * b - 4 * a * c;
if (m > 0)
{
x1 = (-b - Math.Sqrt(b * b - 4 * a * c)) / 2 * a;
x2 = (-b + Math.Sqrt(b * b - 4 * a * c)) / 2 * a;
return Math.Min(x1, x2).ToString();
}
else if (m == 0)
{
x1 = -b / 2 * a;
return x3.ToString();
}
else
return "IMP";
}
You can also use only one condition by combining the first two cases:
private static string GetMinX(int a, int b, int c)
{
double x1, x2, m;
m = b * b - 4 * a * c;
if (m >= 0)
{
x1 = (-b - Math.Sqrt(b * b - 4 * a * c)) / 2 * a;
x2 = (-b + Math.Sqrt(b * b - 4 * a * c)) / 2 * a;
return Math.Min(x1, x2).ToString();
}
else
return "IMP";
}

Is it true equation written on C #?

The equation is:
Is it true I have written?
Double x = 14.26
Double y = -1.22
Double z = 3.5 * Math.Pow(10.0, -2)
Double t;
t = ( 2 * Math.Cos( x - Math.PI / 6 ) ) / ( 0.5 + Math.Pow( Math.Sin( y ), 2 ) ) * ( (1 + Math.Pow( z, 2 ) ) / ( 3 - Math.Pow( z, 2 ) / 5 ) );
Because the result does not coincide with the above desired result -- t = 0,188451240697501, and I need to t deduced 0.564849.

Double x = 14.26, y = -1.22, z = 3.5 * Math.Pow(10.0, -2), t;
t = (2*Math.Cos(x-Math.PI/6))/(0.5+Math.Pow(Math.Sin(y), 2)) * ( 1 + (Math.Pow(z, 2)) / (3-Math.Pow(z, 2)/5));
Console.WriteLine(t); // 0.5648...
You have parenthesis issue, change ((1 + Math.Pow(z, 2)) to (1 + (Math.Pow(z, 2)), the addition of 1 should be calculated after the division.
Also, consider computing the numerators and denominators separately.

I think that one error it is here:
((1 + Math.Pow(z, 2)) / (3-Math.Pow(z, 2)/5))
you should do:
(1 + (Math.Pow(z, 2) / (3-Math.Pow(z, 2)/5)))

Data type that rounds .(9) to 1 automatically?

I'm trying to find solutions to a simple MDAS (i.e. Multiplication, Division...) problem using C#, and while it mostly gets correct solutions, I have problems when the variables add up to x.99999... so I get wrong answers because it doesn't compute to 1.
For example, if I have:
decimal a = 1M;
decimal b = 2M;
decimal c = 6M;
decimal d = 4M;
decimal e = 7M;
decimal f = 8M;
decimal g = 3M;
decimal h = 5M;
decimal i = 9M;
Console.WriteLine(a + 13 * b / c + d + 12 * e - f - 11);
Console.WriteLine(g * h / i);
Console.WriteLine(a + 13 * b / c + d + 12 * e - f - 11 + g * h / i);
Which gives me:
74.33333333333333333333333333
1.6666666666666666666666666667
75.999999999999999999999999997
But I want:
74.33333333333333333333333333
1.6666666666666666666666666667
76
Is there a way that I can always get a precise answer to .(6)+.(3) = 1 without needing to check and modify the values? If not what is the best way to go about it?

The following will round your double value to within a precision of 0.1:
// parameters
double d = -7.9;
double precision = 0.1;
//the conversion
double v = (Math.Abs(Math.Round(d)-d)) < precision ? Math.Round(d) : d;
//output
int i = (int) v;
Console.WriteLine(i);

I went with #tia and Dai's suggestion to use a Rational data type. I got tompazourek's Rationals NuGet package and used it like this:
Rational a = 1;
Rational b = 2;
Rational c = 6;
Rational d = 4;
Rational e = 7;
Rational f = 8;
Rational g = 3;
Rational h = 5;
Rational i = 9;
Console.WriteLine(a + (Rational)13 * b / c + d + (Rational)12 * e - f - (Rational)11);
Console.WriteLine(g * h / i);
Rational toTest = a + (Rational)13 * b / c + d + (Rational)12 * e - f - (Rational)11 + g * h / i;
Console.WriteLine(toTest);
Console.WriteLine(toTest.Equals(76));
Which gives me:
446/6
15/9
4104/54
True

c# math formula output inconsistency

here is my formula:
y = b*exp(-((x-c)^2)/2(d^2)) if x < c, else y = b*exp(-(x-c)^2/2g^2);
this is how i coded it:
for (double x = 0; x <= 7D; x += .01D)
{
b = 6.0410638; c = 2.1344769; d = 0.59183047; g = 0.77384504;
if (x < c)
y = b * Math.Exp(-Math.Pow((x - c), 2)) / (2 * Math.Pow(d, 2D));
else
y = b * Math.Exp(-Math.Pow((x - c), 2)) / (2 * Math.Pow(g, 2D));
qResults.Rows.Add(x, y);
}
the output:
0.0529826172
0.05528786
0.05768187
0.0601675063
0.0627477
0.06542546
0.06820385
0.07108601
0.074075155
0.0771745443
0.08038754
0.08371756
0.08716809
0.0907426849
0.0944449753
0.0982786641
0.102247514
0.106355369
0.110606134
0.115003787
0.119552381
0.124256022
0.1291189
0.13414526
0.139339417
0.144705743
0.150248691
0.155972764
0.16188252
0.167982608
0.1742777
0.180772528
0.187471911
0.1943807
0.201503783
0.2088461
0.2164127
0.224208578
0.232238829
0.2405086
0.24902302
0.257787317
0.2668067
is there a problem with my syntax?
the output should be:
0.009332048
0.009915393
0.010532198
0.011184179
0.011873133
0.012600931
0.013369526
0.014180954
0.015037339
0.015940891
0.016893914
0.017898806
0.01895806
0.020074273
0.021250142
0.022488471
0.023792173
0.025164271
0.026607905
0.028126332
0.029722928
0.031401194
0.033164757
0.035017372
0.036962928
0.039005447
0.041149089
0.043398156
0.045757092
0.048230485
0.050823073
0.053539744
0.05638554
0.059365657
0.062485449
0.065750429
0.069166271
0.072738815
0.076474061
0.08037818
0.084457508
0.088718551
0.093167983
0.097812651
0.102659571
0.107715931
0.112989091
0.118486582
0.124216105
0.130185532
0.136402905
0.142876432
0.149614489
0.156625615
0.163918513
0.171502045
0.17938523
0.187577239
0.196087395
0.204925165
0.21410016
0.223622125
0.233500937
0.2437466
0.254369235
0.265379079
0.276786473
0.288601856
0.30083576
0.313498797
0.326601652
0.340155077
0.354169874
0.368656891
0.383627009
0.39909113
0.415060167
0.431545027
0.448556606
0.466105768
0.484203337
0.502860078
0.522086688
0.541893773
0.562291841
0.583291279
0.604902341
0.627135126
0.649999569
0.673505413
0.697662199
0.722479244
0.747965622
0.774130146
0.800981347
0.828527458
0.856776387
0.885735705
0.915412619
0.945813957
0.976946142
1.008815173
1.041426608
1.074785537
1.108896564
1.143763787
1.179390775
1.21578055

Based on your formula, the code should be:
for (double x = 0; x <= 7D; x += .01D)
{
b = 6.0410638; c = 2.1344769; d = 0.59183047; g = 0.77384504;
if (x < c)
y = b * Math.Exp(-Math.Pow(x - c, 2) / (2 * Math.Pow(d, 2D)));
else
y = b * Math.Exp(-Math.Pow(x - c, 2) / (2 * Math.Pow(g, 2D)));
qResults.Rows.Add(x, y);
}
Basically, the Math.Exp function should be applied to the result of the division, not just to the result of -Math.Pow(x - c, 2)
Also, it's not very clear based on the formula and your code example whether you need to divide by 2 in the exponent computation, then multiply the result by d^2, or actually divide by 2 * d^2 as you are doing in the implementation.