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 = (xp + xq)/2 AND y = (yp + yq)/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 pointsvoid 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 bisectorvoid 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 linespdd 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 pointsdef 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 bisectordef 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 linesdef 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
