点和线段的关系大致可以有下面几

#include <iostream>
#include <string>
#include <cstring>
#include <stdio.h>
#include <cmath>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{int t;cin>>t;while(t--){double a[3],x[4],y[4];int i;for(i=0;i<3;i++) cin>>a[i];for(i=0;i<3;i++) cin>>x[i]>>y[i];double max1=0,max2=0,sum;double r=a[2];for(i=0;i<3;i++){sum=(x[i]-a[0])*(x[i]-a[0])+(y[i]-a[1])*(y[i]-a[1]);sum=sqrt(sum);if(sum>=a[2]) max1=1;if(sum<=a[2]) max2=1;//cout<<sum<<"   "<<a[2]<<' '<<max1<<' '<<max2<<endl;  <strong>//这个是用来判断点是在圆内还是外</strong>}int z=0;if(max1==1&&max2==1) z=1;else if(max1==1&&max2==0) //这一步证明三个点都在圆外{double a1,b1,c1;double s;for(i=0;i<3;i++){a1=(x[i]-a[0])*(x[i]-a[0])+(y[i]-a[1])*(y[i]-a[1]); <em>//计算点到圆心的距离，三条边构成三角形</em>b1=(x[(i+1)%3]-a[0])*(x[(i+1)%3]-a[0])+(y[(i+1)%3]-a[1])*(y[(i+1)%3]-a[1]);c1=(x[i]-x[(i+1)%3])*(x[i]-x[(i+1)%3])+(y[i]-y[(i+1)%3])*(y[i]-y[(i+1)%3]);if(a1+c1>=b1&&b1+c1>=a1) <strong>//判断是否能构成垂线</strong>{//cout<<a1<<' '<<b1<<' '<<c1<<endl;s=c1*a1-((c1+a1-b1)/2*(c1+a1-b1)/2);  //cout<<s<<endl;<strong> //我大中华的秦九韶公式</strong>if(sqrt(s/c1)<=a[2])   z=1; <strong>//找到垂线的长度。</strong>//cout<<i<<"     "<<sqrt(s/c1)<<' '<<z<<' '<<a[2]<<' '<<s-a[2]<<endl;}}}if(z) cout<<"Yes";else cout<<"No";//cout<<' '<<t;cout<<endl;}return 0;
}