给定圆上三点时的圆方程

给定位于圆上的三个坐标(x1, y1) 、 (x2, y2)和(x3, y3) 。任务是找到圆的方程，然后打印圆的中心和半径。
一般形式的圆方程为x² + y² + 2gx + 2fy + c = 0 ，半径形式为(x – h)² + (y -k)² = r² ，其中(h, k)是圆心圆， r是半径。
例子：

Input: x1 = 1, y1 = 0, x2 = -1, y2 = 0, x3 = 0, y3 = 1 
Output: 
Centre = (0, 0) 
Radius = 1 
The equation of the circle is x² + y² = 1.
Input: x1 = 1, y1 = -6, x2 = 2, y2 = 1, x3 = 5, y3 = 2 
Output: 
Centre = (5, -3) 
Radius = 5 
Equation of the circle is x² + y² -10x + 6y + 9 = 0 
 

方法：众所周知，所有三点都在圆上，因此它们将满足圆的方程，将它们放入一般方程中，我们得到三个方程，其中包含三个变量g 、 f和c ，并通过进一步求解我们可以得到值。我们可以推导出获得 g、f 和 c 值的公式为：

Putting coordinates in eqn of circle, we get: 
x1² + y1² + 2gx1 + 2fy1 + c = 0 – (1) 
x2² + y2² + 2gx2 + 2fy2 + c = 0 – (2) 
x3² + y3² + 2gx3 + 2fy3 + c = 0 – (3)
From (1) we get, 2gx1 = -x1² – y1² – 2fy1 – c – (4) 
From (1) we get, c = -x1² – y1² – 2gx1 – 2fy1 – (5) 
From (3) we get, 2fy3 = -x3² – y3² – 2gx3 – c – (6)
Subtracting eqn (2) from eqn (1) we get, 
2g( x1 – x2 ) = ( x2² -x1² ) + ( y2² – y1² ) + 2f( y2 – y1 ) – (A)
Now putting eqn (5) in (6) we get, 
2fy3 = -x3² – y3² – 2gx3 + x1² + y1² + 2gx1 + 2fy1 – (7)
Now putting value of 2g from eqn (A) in (7) we get, 
2f = ( ( x1² – x3² )( x1 – x2 ) +( y1² – y3² )( x1 – x2 ) + ( x2² – x1² )( x1 – x3 ) + ( y2² – y1² )( x1 – x3 ) ) / ( y3 – y1 )( x1 – x2 ) – ( y2 – y1 )( x1 – x3 )
Similarly we can obtain the values of 2g : 
2g = ( ( x1² – x3² )( y1 – x2 ) +( y1² – y3² )( y1 – y2 ) + ( x2² – x1² )( y1 – y3) + ( y2² – y1² )( y1 – y3 ) ) / ( x3 -x1 )( y1 – y2 ) – ( x2 – x1 )( y1 – y3 )
Putting 2g and 2f in eqn (5) we get the value of c and know we had the equation of circle as x² + y² + 2gx + 2fy + c = 0 
 

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the circle on
// which the given three points lie
void findCircle(int x1, int y1, int x2, int y2, int x3, int y3)
{
    int x12 = x1 - x2;
    int x13 = x1 - x3;
 
    int y12 = y1 - y2;
    int y13 = y1 - y3;
 
    int y31 = y3 - y1;
    int y21 = y2 - y1;
 
    int x31 = x3 - x1;
    int x21 = x2 - x1;
 
    // x1^2 - x3^2
    int sx13 = pow(x1, 2) - pow(x3, 2);
 
    // y1^2 - y3^2
    int sy13 = pow(y1, 2) - pow(y3, 2);
 
    int sx21 = pow(x2, 2) - pow(x1, 2);
    int sy21 = pow(y2, 2) - pow(y1, 2);
 
    int f = ((sx13) * (x12)
             + (sy13) * (x12)
             + (sx21) * (x13)
             + (sy21) * (x13))
            / (2 * ((y31) * (x12) - (y21) * (x13)));
    int g = ((sx13) * (y12)
             + (sy13) * (y12)
             + (sx21) * (y13)
             + (sy21) * (y13))
            / (2 * ((x31) * (y12) - (x21) * (y13)));
 
    int c = -pow(x1, 2) - pow(y1, 2) - 2 * g * x1 - 2 * f * y1;
 
    // eqn of circle be x^2 + y^2 + 2*g*x + 2*f*y + c = 0
    // where centre is (h = -g, k = -f) and radius r
    // as r^2 = h^2 + k^2 - c
    int h = -g;
    int k = -f;
    int sqr_of_r = h * h + k * k - c;
 
    // r is the radius
    float r = sqrt(sqr_of_r);
 
    cout << "Centre = (" << h << ", " << k << ")" << endl;
    cout << "Radius = " << r;
}
 
// Driver code
int main()
{
    int x1 = 1, y1 = 1;
    int x2 = 2, y2 = 4;
    int x3 = 5, y3 = 3;
    findCircle(x1, y1, x2, y2, x3, y3);
 
    return 0;
}

Java

// Java implementation of the approach
import java.text.*;
 
class GFG
{
     
// Function to find the circle on
// which the given three points lie
static void findCircle(int x1, int y1,
                        int x2, int y2,
                        int x3, int y3)
{
    int x12 = x1 - x2;
    int x13 = x1 - x3;
 
    int y12 = y1 - y2;
    int y13 = y1 - y3;
 
    int y31 = y3 - y1;
    int y21 = y2 - y1;
 
    int x31 = x3 - x1;
    int x21 = x2 - x1;
 
    // x1^2 - x3^2
    int sx13 = (int)(Math.pow(x1, 2) -
                    Math.pow(x3, 2));
 
    // y1^2 - y3^2
    int sy13 = (int)(Math.pow(y1, 2) -
                    Math.pow(y3, 2));
 
    int sx21 = (int)(Math.pow(x2, 2) -
                    Math.pow(x1, 2));
                     
    int sy21 = (int)(Math.pow(y2, 2) -
                    Math.pow(y1, 2));
 
    int f = ((sx13) * (x12)
            + (sy13) * (x12)
            + (sx21) * (x13)
            + (sy21) * (x13))
            / (2 * ((y31) * (x12) - (y21) * (x13)));
    int g = ((sx13) * (y12)
            + (sy13) * (y12)
            + (sx21) * (y13)
            + (sy21) * (y13))
            / (2 * ((x31) * (y12) - (x21) * (y13)));
 
    int c = -(int)Math.pow(x1, 2) - (int)Math.pow(y1, 2) -
                                2 * g * x1 - 2 * f * y1;
 
    // eqn of circle be x^2 + y^2 + 2*g*x + 2*f*y + c = 0
    // where centre is (h = -g, k = -f) and radius r
    // as r^2 = h^2 + k^2 - c
    int h = -g;
    int k = -f;
    int sqr_of_r = h * h + k * k - c;
 
    // r is the radius
    double r = Math.sqrt(sqr_of_r);
    DecimalFormat df = new DecimalFormat("#.#####");
    System.out.println("Centre = (" + h + "," + k + ")");
    System.out.println("Radius = " + df.format(r));
}
 
// Driver code
public static void main (String[] args)
{
    int x1 = 1, y1 = 1;
    int x2 = 2, y2 = 4;
    int x3 = 5, y3 = 3;
    findCircle(x1, y1, x2, y2, x3, y3);
}
}
 
// This code is contributed by chandan_jnu

Python3

# Python3 implementation of the approach
from math import sqrt
 
# Function to find the circle on
# which the given three points lie
def findCircle(x1, y1, x2, y2, x3, y3) :
    x12 = x1 - x2;
    x13 = x1 - x3;
 
    y12 = y1 - y2;
    y13 = y1 - y3;
 
    y31 = y3 - y1;
    y21 = y2 - y1;
 
    x31 = x3 - x1;
    x21 = x2 - x1;
 
    # x1^2 - x3^2
    sx13 = pow(x1, 2) - pow(x3, 2);
 
    # y1^2 - y3^2
    sy13 = pow(y1, 2) - pow(y3, 2);
 
    sx21 = pow(x2, 2) - pow(x1, 2);
    sy21 = pow(y2, 2) - pow(y1, 2);
 
    f = (((sx13) * (x12) + (sy13) *
          (x12) + (sx21) * (x13) +
          (sy21) * (x13)) // (2 *
          ((y31) * (x12) - (y21) * (x13))));
             
    g = (((sx13) * (y12) + (sy13) * (y12) +
          (sx21) * (y13) + (sy21) * (y13)) //
          (2 * ((x31) * (y12) - (x21) * (y13))));
 
    c = (-pow(x1, 2) - pow(y1, 2) -
         2 * g * x1 - 2 * f * y1);
 
    # eqn of circle be x^2 + y^2 + 2*g*x + 2*f*y + c = 0
    # where centre is (h = -g, k = -f) and
    # radius r as r^2 = h^2 + k^2 - c
    h = -g;
    k = -f;
    sqr_of_r = h * h + k * k - c;
 
    # r is the radius
    r = round(sqrt(sqr_of_r), 5);
 
    print("Centre = (", h, ", ", k, ")");
    print("Radius = ", r);
 
# Driver code
if __name__ == "__main__" :
     
    x1 = 1 ; y1 = 1;
    x2 = 2 ; y2 = 4;
    x3 = 5 ; y3 = 3;
    findCircle(x1, y1, x2, y2, x3, y3);
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to find the circle on
// which the given three points lie
static void findCircle(int x1, int y1,
                        int x2, int y2,
                        int x3, int y3)
{
    int x12 = x1 - x2;
    int x13 = x1 - x3;
 
    int y12 = y1 - y2;
    int y13 = y1 - y3;
 
    int y31 = y3 - y1;
    int y21 = y2 - y1;
 
    int x31 = x3 - x1;
    int x21 = x2 - x1;
 
    // x1^2 - x3^2
    int sx13 = (int)(Math.Pow(x1, 2) -
                    Math.Pow(x3, 2));
 
    // y1^2 - y3^2
    int sy13 = (int)(Math.Pow(y1, 2) -
                    Math.Pow(y3, 2));
 
    int sx21 = (int)(Math.Pow(x2, 2) -
                    Math.Pow(x1, 2));
                     
    int sy21 = (int)(Math.Pow(y2, 2) -
                    Math.Pow(y1, 2));
 
    int f = ((sx13) * (x12)
            + (sy13) * (x12)
            + (sx21) * (x13)
            + (sy21) * (x13))
            / (2 * ((y31) * (x12) - (y21) * (x13)));
    int g = ((sx13) * (y12)
            + (sy13) * (y12)
            + (sx21) * (y13)
            + (sy21) * (y13))
            / (2 * ((x31) * (y12) - (x21) * (y13)));
 
    int c = -(int)Math.Pow(x1, 2) - (int)Math.Pow(y1, 2) -
                                2 * g * x1 - 2 * f * y1;
 
    // eqn of circle be x^2 + y^2 + 2*g*x + 2*f*y + c = 0
    // where centre is (h = -g, k = -f) and radius r
    // as r^2 = h^2 + k^2 - c
    int h = -g;
    int k = -f;
    int sqr_of_r = h * h + k * k - c;
 
    // r is the radius
    double r = Math.Round(Math.Sqrt(sqr_of_r), 5);
 
    Console.WriteLine("Centre = (" + h + "," + k + ")");
    Console.WriteLine("Radius = " + r);
}
 
// Driver code
static void Main()
{
    int x1 = 1, y1 = 1;
    int x2 = 2, y2 = 4;
    int x3 = 5, y3 = 3;
    findCircle(x1, y1, x2, y2, x3, y3);
}
}
 
// This code is contributed by chandan_jnu

Javascript

 
// Function to find the circle on
// which the given three points lie
function findCircle(x1,  y1,  x2,  y2, x3, y3)
{
    var x12 = (x1 - x2);
    var x13 = (x1 - x3);
 
    var y12 =( y1 - y2);
    var y13 = (y1 - y3);
 
    var y31 = (y3 - y1);
    var y21 = (y2 - y1);
 
    var x31 = (x3 - x1);
    var x21 = (x2 - x1);
 
    //x1^2 - x3^2
    var sx13 = Math.pow(x1, 2) - Math.pow(x3, 2);
 
    // y1^2 - y3^2
    var sy13 = Math.pow(y1, 2) - Math.pow(y3, 2);
 
    var sx21 = Math.pow(x2, 2) - Math.pow(x1, 2);
    var sy21 = Math.pow(y2, 2) - Math.pow(y1, 2);
 
    var f = ((sx13) * (x12)
            + (sy13) * (x12)
            + (sx21) * (x13)
            + (sy21) * (x13))
            / (2 * ((y31) * (x12) - (y21) * (x13)));
    var g = ((sx13) * (y12)
            + (sy13) * (y12)
            + (sx21) * (y13)
            + (sy21) * (y13))
            / (2 * ((x31) * (y12) - (x21) * (y13)));
 
    var c = -(Math.pow(x1, 2)) -
    Math.pow(y1, 2) - 2 * g * x1 - 2 * f * y1;
 
    // eqn of circle be
    // x^2 + y^2 + 2*g*x + 2*f*y + c = 0
    // where centre is (h = -g, k = -f) and radius r
    // as r^2 = h^2 + k^2 - c
    var h = -g;
    var k = -f;
    var sqr_of_r = h * h + k * k - c;
 
    // r is the radius
    var r = Math.sqrt(sqr_of_r);
 
    document.write("Centre = (" + h + ", "+ k +")" + "<br>");
    document.write( "Radius = " + r.toFixed(5));
}
 
var x1 = 1, y1 = 1;
    var x2 = 2, y2 = 4;
    var x3 = 5, y3 = 3;
    findCircle(x1, y1, x2, y2, x3, y3);
 
 

输出：

Centre = (3, 2)
Radius = 2.23607

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