ZOJ 1199 Point of Intersection(平面几何)

Point of Intersection

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Given two circles on the same plane which are centered at (x1,y1) and (x2,y2) ,with radiuses r1 and r2, respectively.We can see that they have two common tangent lines in most of the cases.Now you are asked to write a programme to calculate the point of intersection of the two tangents if there exists one. ( See Figure 1 )

Figure. 1 Point of intersection

Input

The input data consists of the information of several figures.The first line of the input contains the number of figures. 
Each figure is described by two lines of data.Each line contains 3 integers constituting the coordinates of the center (x, y) and the radius r (>0) of a circle.

Output

For each figure, you are supposed to output the coordinates (x, y) of the point of intersection if it exists.The x and y must be rounded to two decimal places and be separated by one space.If there is no such point exists simply output "Impossible."

Sample Input

20 0 100 0 50 0 1010 0 1

Output for the Sample Input

Impossible.11.11 0.00

Notice

The common tangent lines like the following figure don't take into account;

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Source: Zhejiang University Local Contest 2002, Preliminary

此题给出两个圆的圆心和半径，如果两圆内含或者两个圆大小相同，则输出impossible.

其余情况根据两个圆心来求出交点坐标

#include<iostream>#include<cstdio>#include<math.h>#include<vector>#include<algorithm>using namespace std;int flag;int b,c;#define eps 1e-6double dis(double x1,double y1,double x2,double y2){return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}int main(){int t;scanf("%d",&t);while(t--){double x1,y1,r1,x2,y2,r2;scanf("%lf%lf%lf%lf%lf%lf",&x1,&y1,&r1,&x2,&y2,&r2);if(r2>r1){swap(r2,r1);swap(x1,x2);swap(y1,y2);}double d1=dis(x1,y1,x2,y2);if(r2+d1-r1<=eps||r1==r2){printf("Impossible.\n");continue;}double d2=d1/(r1/r2-1);if(x1==x2){if(y1<y2)printf("%.2lf %.2lf\n",x2,y2+d2);else printf("%.2lf %.2lf\n",x2,y2-d2);continue;}else{double k=fabs((y1-y2)/(x1-x2));//double b=(x1*y2-x2*y1)/(x1-x2);double by=d2/sqrt(k*k+1)*k;double bx=d2/sqrt(k*k+1);if(y2>=y1&&x2>=x1){printf("%.2lf %.2lf\n",x2+bx,y2+by);}else if(y2<=y1&&x2>=x1){printf("%.2lf %.2lf\n",x2+bx,y2-by);}else if(y2>=y1&&x2<=x1){printf("%.2lf %.2lf\n",x2-bx,y2+by);}else if(y2<=y1&&x2<=x1){printf("%.2lf %.2lf\n",x2-bx,y2-by);}}}return 0;}