Equation of circle when three points on the circle are given

Last Updated : 14 Aug, 2024

Given three coordinates that lie on a circle, (x1, y1), (x2, y2), and (x3, y3). The task is to find the equation of the circle and then print the centre and the radius of the circle.
The equation of circle in general form is x² + y² + 2gx + 2fy + c = 0 and in radius form is (x – h)² + (y -k)² = r², where (h, k) is the centre of the circle and r is the radius.

Examples:

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

Approach: As we know all three-point lie on the circle, so they will satisfy the equation of a circle and by putting them in the general equation we get three equations with three variables g, f, and c, and by further solving we can get the values. We can derive the formula to obtain the value of g, f, and c as:

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 – y2 ) +( 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

Below is the implementation of the above approach:

C++

// C++ implementation of the approach #include <bits/stdc++.h> 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

Python

# 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# 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

<script>  // 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);   </script>

Output

Centre = (3, 2) Radius = 2.23607

Time Complexity: O(log(n))
Auxiliary Space: O(1)

Satisfy the parabola when point (A, B) and the equation is given

gyanendra371

Improve

Article Tags :

Practice Tags :

Equation of circle when three points on the circle are given

Similar Reads