题意：
给一个圆和一个三角形，问圆和三角形是否相交。
思路：
容易发现，三角形中里圆心最远的点一定是三个顶点之一，三角形中里圆心最近的点求点到线段最短距离即可。
当着六个点中有在圆上的，或者有一个在圆内，一个在圆外的，则相交。
代码：
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <sstream>
#define pb push_back
#define X first
#define Y second
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pii pair<int,int>
#define qclear(a) while(!a.empty())a.pop();
#define lowbit(x) (x&-x)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define mst(a,b) memset(a,b,sizeof(a))
#define cout3(x,y,z) cout<<x<<" "<<y<<" "<<z<<endl
#define cout2(x,y) cout<<x<<" "<<y<<endl
#define cout1(x) cout<<x<<endl
#define IOS std::ios::sync_with_stdio(false)
#define SRAND srand((unsigned int)(time(0)))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
using namespace std;
const double PI=acos(-1.0);
const int INF=0x3f3f3f3f;
const ll mod=998244353;
const double eps=1e-8;
const int maxn=2005;
const int maxm=10005;
struct point{
double x,y;
point(){}
point(double _x,double _y){
x=_x;
y=_y;
}
point operator - (const point &b)const{
return point(x-b.x,y-b.y);
}
double operator *(const point &b)const{
return x*b.x+y*b.y;
}
};
struct line{
point s,e;
line(){}
line(point _s,point _e){
s=_s;
e=_e;
}
};
double dist(point a,point b){
return sqrt((a-b)*(a-b));
}
point getnear(point p,line l){
point res;
double t=((p-l.s)*(l.e-l.s))/((l.e-l.s)*(l.e-l.s));
if(t>=0&&t<=1){
res.x=l.s.x+(l.e.x-l.s.x)*t;
res.y=l.s.y+(l.e.y-l.s.y)*t;
}else{
if(dist(p,l.s)<dist(p,l.e))res=l.s;
else res=l.e;
}
return res;
}
point p[5];
line lines[5];
void solve() {
int t;
cin>>t;
while(t--){
int e=0,l=0,g=0;
bool ok=0,out=0,in=0;
point c;
double r;
cin>>c.x>>c.y>>r;
for(int i=0;i<3;i++){
cin>>p[i].x>>p[i].y;
double nowdis=dist(p[i],c);
if(fabs(nowdis-r)<eps){
e++;
}else if(nowdis<r){
l++;
}else{
g++;
}
}
if(e){
ok=1;
}
if(g){
out=1;
}
l=0;e=0;g=0;
lines[0].s=p[0];
lines[0].e=p[1];
lines[1].s=p[0];
lines[1].e=p[2];
lines[2].s=p[1];
lines[2].e=p[2];
for(int i=0;i<3;i++){
double nowdis=dist(getnear(c,lines[i]),c);
if(fabs(nowdis-r)<eps){
e++;
}else if(nowdis<r){
l++;
}else{
g++;
}
}
if(e){
ok=1;
}
if(l){
in=1;
}
if(ok||(out&&in)){
cout<<"Yes"<<endl;
}else{
cout<<"No"<<endl;
}
}
return ;
}
int main() {
#ifdef LOCAL
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#else
// freopen("","r",stdin);
// freopen("","w",stdout);
#endif
solve();
return 0;
}