Given 3 non-collinear points in the 2D Plane P, Q and R with their respective x and y coordinates, find the circumcenter of the triangle.**Note:** Circumcenter of a triangle is the centre of the circle, formed by the three vertices of a triangle. Note that three points can uniquely determine a circle.

Examples:

Input : P(6, 0) Q(0, 0) R(0, 8) Output : The circumcenter of the triangle PQR is: (3, 4) Input : P(1, 1) Q(0, 0) R(2, 2) Output : The two perpendicular bisectors found come parallel. Thus, the given points do not form a triangle and are collinear

Given, three points of the triangle, we can easily find the sides of the triangle. Now, we have the equations of the lines for the three sides of the triangle. After getting these, we can find the circumcenter of the triangle by a simple property stated as under:

The circumcenter of the triangle is point where all the perpendicular bisectors of the sides of the triangle intersect.

This is well explained in the following diagram.

Note here that, there is no need to find all of the three sides of the triangle. Finding two sides is sufficient as we can uniquely find the point of intersection using just two perpendicular bisectors. The third perpendicular bisector will itself pass through the so found circumcenter.

The things to be done can be divided as under:

- Find 2 lines (say PQ and QR) which form the sides of the triangle.
- Find the perpendicular bisectors of PQ and QR (say lines L and M respectively).
- Find the point of intersection of lines L and M as the circumcenter of the given triangle.

**STEP 1**

Refer this post Program to find line passing through 2 Points**STEP 2**

Let PQ be represented as ax + by = c

A line perpendicular to this line is represented as -bx + ay = d for some d.

However, we are interested in the perpendicular bisector. So, we find the mid-point of P and Q and putting this value in the standard equation, we get the value of d.

Similarly, we repeat the process for QR.

d = -bx + ay where, x = (x_{p}+ x_{q})/2 AND y = (y_{p}+ y_{q})/2

**STEP 3**

Refer this post Program for Point of Intersection of Two Lines

## CPP

`// C++ program to find the CIRCUMCENTER of a` `// triangle` `#include <iostream>` `#include <cfloat>` `using` `namespace` `std;` `// This pair is used to store the X and Y` `// coordinate of a point respectively` `#define pdd pair<double, double>` `// Function to find the line given two points` `void` `lineFromPoints(pdd P, pdd Q, ` `double` `&a,` ` ` `double` `&b, ` `double` `&c)` `{` ` ` `a = Q.second - P.second;` ` ` `b = P.first - Q.first;` ` ` `c = a*(P.first)+ b*(P.second);` `}` `// Function which converts the input line to its` `// perpendicular bisector. It also inputs the points` `// whose mid-point lies on the bisector` `void` `perpendicularBisectorFromLine(pdd P, pdd Q,` ` ` `double` `&a, ` `double` `&b, ` `double` `&c)` `{` ` ` `pdd mid_point = make_pair((P.first + Q.first)/2,` ` ` `(P.second + Q.second)/2);` ` ` `// c = -bx + ay` ` ` `c = -b*(mid_point.first) + a*(mid_point.second);` ` ` `double` `temp = a;` ` ` `a = -b;` ` ` `b = temp;` `}` `// Returns the intersection point of two lines` `pdd lineLineIntersection(` `double` `a1, ` `double` `b1, ` `double` `c1,` ` ` `double` `a2, ` `double` `b2, ` `double` `c2)` `{` ` ` `double` `determinant = a1*b2 - a2*b1;` ` ` `if` `(determinant == 0)` ` ` `{` ` ` `// The lines are parallel. This is simplified` ` ` `// by returning a pair of FLT_MAX` ` ` `return` `make_pair(FLT_MAX, FLT_MAX);` ` ` `}` ` ` `else` ` ` `{` ` ` `double` `x = (b2*c1 - b1*c2)/determinant;` ` ` `double` `y = (a1*c2 - a2*c1)/determinant;` ` ` `return` `make_pair(x, y);` ` ` `}` `}` `void` `findCircumCenter(pdd P, pdd Q, pdd R)` `{` ` ` `// Line PQ is represented as ax + by = c` ` ` `double` `a, b, c;` ` ` `lineFromPoints(P, Q, a, b, c);` ` ` `// Line QR is represented as ex + fy = g` ` ` `double` `e, f, g;` ` ` `lineFromPoints(Q, R, e, f, g);` ` ` `// Converting lines PQ and QR to perpendicular` ` ` `// vbisectors. After this, L = ax + by = c` ` ` `// M = ex + fy = g` ` ` `perpendicularBisectorFromLine(P, Q, a, b, c);` ` ` `perpendicularBisectorFromLine(Q, R, e, f, g);` ` ` `// The point of intersection of L and M gives` ` ` `// the circumcenter` ` ` `pdd circumcenter =` ` ` `lineLineIntersection(a, b, c, e, f, g);` ` ` `if` `(circumcenter.first == FLT_MAX &&` ` ` `circumcenter.second == FLT_MAX)` ` ` `{` ` ` `cout << ` `"The two perpendicular bisectors "` ` ` `"found come parallel"` `<< endl;` ` ` `cout << ` `"Thus, the given points do not form "` ` ` `"a triangle and are collinear"` `<< endl;` ` ` `}` ` ` `else` ` ` `{` ` ` `cout << ` `"The circumcenter of the triangle PQR is: "` `;` ` ` `cout << ` `"("` `<< circumcenter.first << ` `", "` ` ` `<< circumcenter.second << ` `")"` `<< endl;` ` ` `}` `}` `// Driver code.` `int` `main()` `{` ` ` `pdd P = make_pair(6, 0);` ` ` `pdd Q = make_pair(0, 0);` ` ` `pdd R = make_pair(0, 8);` ` ` `findCircumCenter(P, Q, R);` ` ` `return` `0;` `}` |

## Python3

`# Python3 program to find the CIRCUMCENTER of a` `# triangle` `# This pair is used to store the X and Y` `# coordinate of a point respectively` `# define pair<double, double>` `# Function to find the line given two points` `def` `lineFromPoints(P, Q, a, b, c):` ` ` `a ` `=` `Q[` `1` `] ` `-` `P[` `1` `]` ` ` `b ` `=` `P[` `0` `] ` `-` `Q[` `0` `]` ` ` `c ` `=` `a ` `*` `(P[` `0` `]) ` `+` `b ` `*` `(P[` `1` `])` ` ` `return` `a, b, c` `# Function which converts the input line to its` `# perpendicular bisector. It also inputs the points` `# whose mid-point lies on the bisector` `def` `perpendicularBisectorFromLine(P, Q, a, b, c):` ` ` `mid_point ` `=` `[(P[` `0` `] ` `+` `Q[` `0` `])` `/` `/` `2` `, (P[` `1` `] ` `+` `Q[` `1` `])` `/` `/` `2` `]` ` ` `# c = -bx + ay` ` ` `c ` `=` `-` `b ` `*` `(mid_point[` `0` `]) ` `+` `a ` `*` `(mid_point[` `1` `])` ` ` `temp ` `=` `a` ` ` `a ` `=` `-` `b` ` ` `b ` `=` `temp` ` ` `return` `a, b, c` `# Returns the intersection point of two lines` `def` `lineLineIntersection(a1, b1, c1, a2, b2, c2):` ` ` `determinant ` `=` `a1 ` `*` `b2 ` `-` `a2 ` `*` `b1` ` ` `if` `(determinant ` `=` `=` `0` `):` ` ` ` ` `# The lines are parallel. This is simplified` ` ` `# by returning a pair of (10.0)**19` ` ` `return` `[(` `10.0` `)` `*` `*` `19` `, (` `10.0` `)` `*` `*` `19` `]` ` ` `else` `:` ` ` `x ` `=` `(b2 ` `*` `c1 ` `-` `b1 ` `*` `c2)` `/` `/` `determinant` ` ` `y ` `=` `(a1 ` `*` `c2 ` `-` `a2 ` `*` `c1)` `/` `/` `determinant` ` ` `return` `[x, y]` `def` `findCircumCenter(P, Q, R):` ` ` ` ` `# Line PQ is represented as ax + by = c` ` ` `a, b, c ` `=` `0.0` `, ` `0.0` `, ` `0.0` ` ` `a, b, c ` `=` `lineFromPoints(P, Q, a, b, c)` ` ` `# Line QR is represented as ex + fy = g` ` ` `e, f, g ` `=` `0.0` `, ` `0.0` `, ` `0.0` ` ` `e, f, g ` `=` `lineFromPoints(Q, R, e, f, g)` ` ` `# Converting lines PQ and QR to perpendicular` ` ` `# vbisectors. After this, L = ax + by = c` ` ` `# M = ex + fy = g` ` ` `a, b, c ` `=` `perpendicularBisectorFromLine(P, Q, a, b, c)` ` ` `e, f, g ` `=` `perpendicularBisectorFromLine(Q, R, e, f, g)` ` ` `# The point of intersection of L and M gives` ` ` `# the circumcenter` ` ` `circumcenter ` `=` `lineLineIntersection(a, b, c, e, f, g)` ` ` `if` `(circumcenter[` `0` `] ` `=` `=` `(` `10.0` `)` `*` `*` `19` `and` `circumcenter[` `1` `] ` `=` `=` `(` `10.0` `)` `*` `*` `19` `):` ` ` `print` `(` `"The two perpendicular bisectors found come parallel"` `)` ` ` `print` `(` `"Thus, the given points do not form a triangle and are collinear"` `)` ` ` `else` `:` ` ` `print` `(` `"The circumcenter of the triangle PQR is: "` `, end` `=` `"")` ` ` `print` `(` `"("` `, circumcenter[` `0` `], ` `","` `, circumcenter[` `1` `], ` `")"` `)` `# Driver code.` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `P ` `=` `[` `6` `, ` `0` `]` ` ` `Q ` `=` `[` `0` `, ` `0` `]` ` ` `R ` `=` `[` `0` `, ` `8` `]` ` ` `findCircumCenter(P, Q, R)` ` ` `# This code is contributed by mohit kumar 29` |

Output:

The circumcenter of the triangle PQR is: (3, 4)

This article is contributed by **Aanya Jindal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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