1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 using namespace std;
 6 
 7 const double eps=1e-8;
 8 const double Pi=acos(-1.0);
 9 struct Point
10 {
11     double x,y;
12     Point(double x=0,double y=0):x(x),y(y) {}
13 };
14 typedef Point Vector;
15 Vector operator +(Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
16 Vector operator -(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}
17 Vector operator *(Vector A,double p){return Vector(A.x*p,A.y*p);}
18 Vector operator /(Vector A,double p){return Vector(A.x/p,A.y/p);}
19 double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}//点积
20 double Length(Vector A){return sqrt(Dot(A,A));}//向量的长度
21 inline double min(double a,double b){return a>b?b:a;}
22 const int maxn=30;
23 int n;
24 struct circle
25 {
26     Point c;
27     double r;
28 }C[maxn];
29 
30 double getarea(int id,int i,double r)
31 {
32     double  clen=Length(C[id].c-C[i].c);
33     if(clen>=r+C[i].r) return 0;
34     double t=min(r,C[i].r);
35     if(clen<=fabs(r-C[i].r)) return Pi*t*t;
36     double ang1=acos((r*r+clen*clen-C[i].r*C[i].r)/(2*r*clen));
37     double ang2=acos((C[i].r*C[i].r+clen*clen-r*r)/(2*C[i].r*clen));
38     double area=ang1*r*r+ang2*C[i].r*C[i].r;
39     area-=sin(ang2)*C[i].r*clen;
40     return area;
41 }
42 
43 bool judge(int id,double r)
44 {
45     for(int i=0;i<n;i++)
46     {
47         double area=getarea(id,i,r);
48         if(Pi*C[i].r*C[i].r>2*area)
49             return false;
50     }
51     return true;
52 }
53 
54 double binary_search(double l,double r,int id)
55 {
56     double mid;
57     while(r-l>eps)
58     {
59         mid=(l+r)/2.0;
60         if(judge(id,mid)) r=mid;
61         else l=mid;
62     }
63     return l;
64 }
65 
66 void solve()
67 {
68     double ans=50000;
69     for(int i=0;i<n;i++)
70         ans=min(ans,binary_search(0,40000,i));
71     printf("%.4lf\n",ans);
72 }
73 int main()
74 {
75     int i,t;
76     scanf("%d",&t);
77     while(t--)
78     {    
79         scanf("%d",&n);
80         for(i=0;i<n;i++)
81             scanf("%lf%lf%lf",&C[i].c.x,&C[i].c.y,&C[i].r);
82         solve();
83     }
84     return 0;
85 }