#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct P
{
double x,y;
};
struct circle
{
P c;
double r;
};
int cmp(double x)//误差处理
{
if(fabs(x)<1e-15) return 0;
if(x>0) return 1;
return -1;
}
bool judge(double a,double b,double c,double &x1,double &x2)//一元二次方程求解
{
double tep=b*b-4*a*c;
if(cmp(tep)<0) return 0;
x1=(-b+sqrt(tep))/(2*a);
x2=(-b-sqrt(tep))/(2*a);
return 1;
}
bool judge(circle o,P a,P c) //线段与圆的关系
{
double k,b;
if(cmp(a.x-c.x)==0)//直线AC垂直与X轴(斜率不存在)
{
double t=o.r*o.r-(a.x-o.c.x)*(a.x-o.c.x);
if(t<0) return 0;//圆心到直线的距离大于半径,不相交
double maxy=max(a.y,c.y),miny=min(a.y,c.y),y1=sqrt(t)+o.c.y,y2=o.c.y-sqrt(t);//分别表示点A、C中纵坐标最小、最大的值,以及直线AC与圆的两个交点的纵坐标。
if(cmp(miny-y1)<=0&&cmp(y1-maxy)<=0) return 1;//AC分居上交点两侧(线段与圆有交点)
if(cmp(miny-y2)<=0&&cmp(y2-maxy)<=0) return 1;//AC分居下交点两侧(线段与圆有交点)
return 0;
}
//斜率存在时,斜率=k
k=(a.y-c.y)/(a.x-c.x);
b=a.y-k*a.x;
double x1,x2;
int f=judge(k*k+1,2*k*(b-o.c.y)-2*o.c.x,o.c.x*o.c.x+(b-o.c.y)*(b-o.c.y)-o.r*o.r,x1,x2);//利用解析几何知识求出直线AC与圆的交点的横坐标
if(f==0) return 0;//无交点
int maxx=max(a.x,c.x),minx=min(a.x,c.x);//同上
if(cmp(minx-x1)<=0&&cmp(x1-maxx)<=0) return 1;
if(cmp(minx-x2)<=0&&cmp(x2-maxx)<=0) return 1;
return 0;
}
bool judge(circle o,P a,P b,P c)
{
if(judge(o,a,b)||judge(o,a,c)||judge(o,b,c)) return 1;
return 0;
}
int main()
{
int t;
circle o;
P a,b,c;
scanf("%d",&t);
while(t--)
{
scanf("%lf%lf%lf",&o.c.x,&o.c.y,&o.r);
scanf("%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);
if(judge(o,a,b,c)) puts("Yes");
else puts("No");
}
return 0;
}