**UVA Problem 438 – The Circumference of the Circle Solution:**

Click here to go to this problem in uva Online Judge.

**Solving Technique:**

Given three points find the circumference of the circle. Points are non-collinear meaning they are not on a straight line.

Have a look at Circumscribed circle, Semi perimeter and Heron’s formula to find more about the formula and derivation.

Also related to this have a look at UVA 10432 – Polygon Inside a Circle.

** Important: **Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.

More Inputs of This Problem on uDebug.

**Input:**

0.0 -0.5 0.5 0.0 0.0 0.5 0.0 0.0 0.0 1.0 1.0 1.0 5.0 5.0 5.0 7.0 4.0 6.0 0.0 0.0 -1.0 7.0 7.0 7.0 50.0 50.0 50.0 70.0 40.0 60.0 0.0 0.0 10.0 0.0 20.0 1.0 0.0 -500000.0 500000.0 0.0 0.0 500000.0

**Output:**

3.14 4.44 6.28 31.42 62.83 632.24 3141592.65

### Code:

/** * Author: Asif Ahmed * Site: https://quickgrid.wordpress.com * Problem: UVA 438 - The Circumference of the Circle. * Technique: Finding diameter and circumference of Circumscribed * circle or Circumcircle. * Using Heron's formula to calculate semi perimeter * and area of triangle from Three points. */ #include<stdio.h> #include<string.h> #include<math.h> #define PI 3.141592653589793 int main(){ //freopen("input.txt", "r", stdin); //freopen("output.txt", "w", stdout); double x1, y1; double x2, y2; double x3, y3; while( scanf("%lf%lf%lf%lf%lf%lf", &x1, &y1, &x2, &y2, &x3, &y3 ) == 6 ){ // a dist between (x1,y1) and (x2,y2) // b dist between (x2,y2) and (x3,y3) // c dist between (x3,y3) and (x1,y1) double a = sqrt( pow(x1 - x2, 2) + pow(y1 - y2, 2) ); double b = sqrt( pow(x2 - x3, 2) + pow(y2 - y3, 2) ); double c = sqrt( pow(x3 - x1, 2) + pow(y3 - y1, 2) ); // semi perimeter, s = (a+b+c)/2 double s = ( a + b + c ) / 2; // Area using Heron's Formula double A = sqrt( s*(s-a)*(s-b)*(s-c) ); // Diameter of circumscribed circle d = abc/2A double d = (a * b * c) / (2 * A); // Result circumference of the circumcircle or circumscribed circle double circumference = PI * d; printf("%.2lf\n", circumference ); } return 0; }

### Unnecessary Complex Code:

/** * Author: Asif Ahmed * Site: https://quickgrid.wordpress.com * Problem: UVA 438 - The Circumference of the Circle. * Technique: Point and Line representation in structure. * Finding diameter and circumference of Circumscribed * circle or Circumcircle. * Using Heron's formula to calculate semi perimeter * and area of triangle from Three points. */ #include<stdio.h> #include<string.h> #include<math.h> #define PI 3.141592653589793 struct Point{ double x; double y; }; struct Line{ Point a; Point b; Point setPoints( Point _a = {0.0,0.0}, Point _b = {0.0,0.0} ){ a = _a; b = _b; } double length(){ return sqrt( pow(a.x - b.x, 2) + pow(a.y - b.y, 2) ); } }; int main(){ //freopen("input.txt", "r", stdin); //freopen("output.txt", "w", stdout); Point point[3]; Line line [3]; while( scanf("%lf%lf%lf%lf%lf%lf", &point[0].x, &point[0].y, &point[1].x, &point[1].y, &point[2].x, &point[2].y ) == 6 ){ // Loop is same as code below. //line[0].setPoints(point[0], point[1]); //line[1].setPoints(point[1], point[2]); //line[2].setPoints(point[2], point[0]); int N = 3; for( int i = 0; i < N; ++i ) line[i].setPoints( point[i], point[(i+1) % N] ); // semi perimeter is half of perimeter. // Perimeter is distance around the shape in 2D. double s = ( line[0].length() + line[1].length() + line[2].length() ) / 2; // Area using Heron's Formula double A = sqrt( s * (s - line[0].length()) * (s - line[1].length()) * (s - line[2].length()) ); // Diameter of circumscribed circle d = abc/2A double d = (line[0].length() * line[1].length() * line[2].length()) / (2 * A); // Result circumference of the circumcircle double circumference = PI * d; printf("%.2lf\n", circumference ); } return 0; }