190-Circle Through Three Points

[Rizoan toufiq]

i got WA.....But all given above input and output is satisfy..... help me guru

my code:

Code: Select all


#include<stdio.h>
#include<math.h>
int main(){
  //freopen("a.in","r",stdin);
  //freopen("b.in","w",stdout);
  double x1,y1,x2,y2,y3,x3;
  double h,k,r,g,f,c;
  int flag=0;
  while(1){
    if(scanf("%lf %lf %lf %lf %lf %lf",&x1,&y1,&x2,&y2,&x3,&y3)!=6)
       break;
   
    double ma,mb;   
    char p='-',q='-',a='+',b='+',s='+';
   
     if(x1==x2)
       ma=1;
    else
       ma = (y1-y2)/(x1-x2);
    if(x2==x3)
       mb=1;
    else
       mb = (y2-y3)/(x2-x3);
   
    h=((y1-y3)*ma*mb + mb*(x1+x2)-ma*(x2+x3))/(2*(mb-ma));
    k=((x3-x1)-(y1+y2)*ma+(y2+y3)*mb)/(2*(mb-ma));
    r=sqrt((x1-h)*(x1-h) + (y1-k)*(y1-k));
   
    g=-2*h;
    f=-2*k;
    //c=h*h+k*k-r*r;
    c = -(x2 * x2 + y2 * y2 + g * x2 + f * y2);

    if(h<=0){
      p='+';
      h=-1*h;
    }
    if(k<=0){
      q ='+';
      k=-1*k;
    }
     printf("(x %c %.3lf)^2 + (y %c %.3lf)^2 = %.3lf^2\n",p,h,q,k,r);
     if(g<0){
       a='-';
      g=-1*g;
    }
    if(f<0){
      b ='-';
      f=-1*f;
    }
    if(c<0){
       s='-';
      c=-1*c;
    }
    printf("x^2 + y^2 %c %.3lfx %c %.3lfy %c %.3lf = 0\n\n",a,g,b,f,s,c);
  }
  return 0;
}



[drcoolsanta]

EDIT: I got an AC on mine... A few pointers though:
1. Use fabs... And not -h
2. No need to make it x^2 rather than (x + 0.000)^2
3. Division by zero can be avoided if you don't find slope.

Aim is to find equation: ax + by + c = 0
Slope:
m = (y2 - y1)/(x2 - x1)
also m = -a/b
take a = y1 - y2
and b = x2 - x1
then c = -(ax1 + by1)
Perpendicular to ax + by + c = 0 is bx - ay + k = 0

Two methods that you could be used:
1. Find two lines passing through the points, then perpendicular bisectors, center and then the radius.
2. Find one line (with first two points), and a random circle passing through them. Find family of circles, third point lies on it.