UVALIVE6457（2013年杭州现场赛D题）

几何题，虽然事实上大部分都是在解各种方程……

题目意思是，给定平面中两个相离的圆和一个在这两个圆外部的点，求所有与两个圆都外切而经过这个点的圆。

首先化简，将坐标原点平移至第三个点，接着可以列出新圆圆心(x,y)与半径r的关系式三条，接着就是去解这个方程。解的过程写起来有点麻烦，忽略了直接看程序好了……

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<vector>
#include<iostream>
using namespace std;
typedef long long ll;

const double eps=1e-9;
int dcmp(double x){if(x<-eps) return -1;return x>eps;}

struct point
{
	double x,y;
	point(double x=0,double y=0):x(x),y(y){}
	void read(){int xt,yt;scanf("%d %d",&xt,&yt);x=xt;y=yt;}
	void output(){printf("%.6lf %.6lf ",x,y);}
};

point operator +(point a,point b){return point(a.x+b.x,a.y+b.y);}
point operator -(point a,point b){return point(a.x-b.x,a.y-b.y);}
point operator *(point a,double b){return point(a.x*b,a.y*b);}
double cross(point a,point b){return a.x*b.y-a.y*b.x;}
double dot(point a,point b){return a.x*b.x+a.y*b.y;}
double length(point a){return sqrt(dot(a,a));}

struct circle
{
	point c;double r;
	circle(point c=point(),double r=0):c(c),r(r){}
	point Point(double a){return point(c.x+cos(a)*r,c.y+sin(a)*r);}
	void read(){int rt;c.read();scanf("%d",&rt);r=rt;}
};

int getlcint(point p,point v,circle C,vector<point>&sol)
{
	sol.clear();
	double a=v.x,b=p.x-C.c.x,c=v.y,d=p.y-C.c.y;
	double e=a*a+c*c,f=2*(a*b+c*d),g=b*b+d*d-C.r*C.r;
	double delta=f*f-4*e*g;
	if(dcmp(delta)<0) return 0;
	if(dcmp(delta)==0) {sol.push_back(p+v*(-f/(2*e)));return 1;}
	sol.push_back(p+v*((-f-sqrt(delta)/(2*e))));sol.push_back(p+v*((-f+sqrt(delta)/(2*e))));return 2;
}

circle c1,c2;
point p3;

pair<double,double> sol12(point t1)
{
	double tt=2*cross(c1.c,c2.c),tt1=t1.x*c2.c.y-t1.y*c1.c.y,tt2=c1.c.x*t1.y-c2.c.x*t1.x;
	pair<double,double> ans(tt1/tt,tt2/tt);
	return ans;
}

int sol2(double A,double B,double C,vector<double> &sol)
{
	sol.clear();
	if(dcmp(A)==0)
	{
		if(dcmp(B)==0) return 0;
		sol.push_back(-C/B);
		return 1;
	}
	double delta=B*B-4*A*C;
	if(dcmp(delta)<0) return 0;
	if(dcmp(delta)==0) {if(dcmp((-B)/(2*A))<=0) return 0;sol.push_back((-B)/(2*A));return 1;}
	if(dcmp((-B-sqrt(delta))/(2*A))>0) sol.push_back((-B-sqrt(delta))/(2*A));
	if(dcmp((-B+sqrt(delta))/(2*A))>0) sol.push_back((-B+sqrt(delta))/(2*A));
	return int(sol.size());
}

void solve()
{
	int i,j;
	if(dcmp(cross(c1.c,c2.c)))
	{
		pair<double,double> ttt1=sol12(point(-2*c1.r,-2*c2.r)),ttt2=sol12(point(dot(c1.c,c1.c)-c1.r*c1.r,dot(c2.c,c2.c)-c2.r*c2.r));
		double a1=ttt1.first,a2=ttt1.second,b1=ttt2.first,b2=ttt2.second;
		double A=a1*a1+a2*a2-1,B=2*(a1*b1+a2*b2),C=b1*b1+b2*b2;
		vector<double> sol;
		int m=sol2(A,B,C,sol);
		printf("%d\n",m);
		for(i=0;i<m;i++) {(p3+point(a1*sol[i]+b1,a2*sol[i]+b2)).output();printf("%.6lf\n",sol[i]);}
		return;
	}
	else
	{
		if(dcmp(c1.c.x*c2.r-c2.c.x*c1.r)==0&&dcmp(c1.c.y*c2.r-c2.c.y*c1.r)==0)
		{
			printf("0\n");return;
		}
		else if(dcmp(c1.c.x*c2.r-c2.c.x*c1.r))
		{
			double r=(c1.c.x*(dot(c2.c,c2.c)-c2.r*c2.r)-c2.c.x*(dot(c1.c,c1.c)-c1.r*c1.r))/(2*(c1.c.x*c2.r-c2.c.x*c1.r));
			if(dcmp(r)<=0) {printf("0\n");return;}
			vector<point> sol;
			double cr=-2*c1.r*r+(dot(c1.c,c1.c)-c1.r*c1.r);
			point yz;if(dcmp(c1.c.x)) yz=point(cr/(2*c1.c.x),0);else yz=point(0,cr/(2*c1.c.y));
			int m=getlcint(yz,point(-2*c1.c.y,2*c1.c.x),circle(point(),r),sol);
			printf("%d\n",m);
			for(i=0;i<m;i++) {(p3+sol[i]).output();printf("%.6lf\n",r);}
		}
		else
		{
			double r=(c1.c.y*(dot(c2.c,c2.c)-c2.r*c2.r)-c2.c.y*(dot(c1.c,c1.c)-c1.r*c1.r))/(2*(c1.c.y*c2.r-c2.c.y*c1.r));
			if(dcmp(r)<=0) {printf("0\n");return;}
			vector<point> sol;
			double cr=-2*c1.r*r+(dot(c1.c,c1.c)-c1.r*c1.r);
			point yz;if(dcmp(c1.c.x)) yz=point(cr/(2*c1.c.x),0);else yz=point(0,cr/(2*c1.c.y));
			int m=getlcint(yz,point(-2*c1.c.y,2*c1.c.x),circle(point(),r),sol);
			printf("%d\n",m);
			for(i=0;i<m;i++) {(p3+sol[i]).output();printf("%.6lf\n",r);}
		}
	}
}

int main()
{
	int i,j,T;
	scanf("%d",&T);
	while(T--)
	{
		c1.read();c2.read();p3.read();
		c1.c=c1.c-p3;c2.c=c2.c-p3;
		solve();
	}
	return 0;
}