ZOJ1199-Point of Intersection

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

2
0 0 10
0 0 5
0 0 10
10 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

题意：求两圆切线相交点的坐标

解题思路：当半径相同，两圆平行，所以不可以；当圆形距小于半径差，有一圆在另一圆内，所以也不可以

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const LL INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;

int main()
{
	int t;
	double x1, x2, y1, y2, r1, r2;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%lf%lf%lf", &x1, &y1, &r1);
		scanf("%lf%lf%lf", &x2, &y2, &r2);
		double dis = sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));
		if (r1 == r2 || dis <= fabs(r1 - r2))
		{
			printf("Impossible.\n");
			continue;
		}
		double x;
		x = r2*dis*1.0 / (r1 - r2);
		double x3 = (x2 - x1)*x / dis + x2;
		double y3 = (y2 - y1)*x / dis + y2;
		printf("%.2lf %.2lf\n", x3, y3);
	}
	return 0;
}