一道解析几何么，，，
其实就是求直线与圆的切线。
看到方法有很多，比如根据角度之类的。
这里主要用到了初中的几何知识。
考虑这幅图。
首先可以根据相似三角形知道b的长度，同时圆心与点的方向也知道。 那么 圆心+b 就是 切点连线 与 点与圆心 连线的交点了。
然后根据 面积，有 l·r = (b的长度)*(中间点到切点的长度) .
就很容易得到切点了。详细看代码，poj返回vector好像会RE，就改成pair了。
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <cstring>
using namespace std;
typedef double db;
const db eps = 1e-;
const db pi = acos(-);
int sign(db k){
if (k>eps) return ; else if (k<-eps) return -; return ;
}
int cmp(db k1,db k2){return sign(k1-k2);}
struct point{
db x,y;
point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
point operator * (db k1) const{return (point){x*k1,y*k1};}
point operator / (db k1) const{return (point){x/k1,y/k1};}
point turn(db k1){ return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
point turn90(){ return (point){-y,x};}
db abs(){ return sqrt(x*x+y*y);}
db abs2(){ return x*x+y*y;}
db dis(point k1){ return (*this-k1).abs();}
point unit(){db w=abs(); return (point){x/w,y/w};}
};
db cross(point k1,point k2){ return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){ return k1.x*k2.x+k1.y*k2.y;}
point proj(point k1,point k2,point q){
point k=k2-k1;
return k1+k*(dot(q-k1,k)/k.abs2());
}
point getLL(point k1,point k2,point k3,point k4){
db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3);
return (k1*w2+k2*w1)/(w1+w2);
}
struct circle{
point o;db r;
int inside(point k){ return cmp(r,o.dis(k));}
};
pair<point,point> TangentCP(circle k1,point k2){//
db a=(k2-k1.o).abs(),b=k1.r*k1.r/a,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));
point k=(k2-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;
return {m-del,m+del};
}
struct line{
db l,r;
};
bool cmp2(line a,line b){
return a.l<b.l;
}
int n;
line l[];
circle c[];
pair<point,point> g;
point e,s1,s2;
int main(){
bool f=;
while (scanf("%d",&n)&&n){
if(f)printf("\n");
f=;
scanf("%lf%lf",&e.x,&e.y);
for(int i=;i<=n;i++){
scanf("%lf%lf%lf",&c[i].o.x,&c[i].o.y,&c[i].r);
}
for(int i=;i<=n;i++){
g=TangentCP(c[i],e);
s1 = getLL(e,g.first,point{0.0,0.0},point{100.0,0.0});
s2 = getLL(e,g.second,point{0.0,0.0},point{100.0,0.0});
l[i]=line{s2.x,s1.x};
}
sort(l+,l++n,cmp2);
db L = l[].l,R = l[].r;
for(int i=;i<=n;i++){
if(l[i].l>R){
printf("%.2f %.2f\n",L,R);
L = l[i].l,R=l[i].r;
} else
R = max(R,l[i].r);
}
printf("%.2f %.2f\n",L,R);
}
}
/**
*
1
300 300
390 150 90
0 6
300 450
70 50 30
120 20 20
270 40 10
250 85 20
220 30 30
380 100 100
*/