（第六届福建省赛）B

 Problem Description

Two different circles can have at most four common tangents.

The picture below is an illustration of two circles with four common tangents.

Now given the center and radius of two circles, your job is to find how many common tangents between them.

 Input

The first line contains an integer T, meaning the number of the cases (1 <= T <= 50.).

For each test case, there is one line contains six integers x1 (−100 ≤ x1 ≤ 100), y1 (−100 ≤ y1 ≤ 100), r1 (0 < r1 ≤ 200), x2 (−100 ≤ x2 ≤ 100), y2 (−100 ≤ y2 ≤ 100), r2 (0 < r2 ≤ 200). Here (x1, y1) and (x2, y2) are the coordinates of the center of the first circle and second circle respectively, r1 is the radius of the first circle and r2 is the radius of the second circle.

 Output

For each test case, output the corresponding answer in one line.

If there is infinite number of tangents between the two circles then output -1.

题意：求两个圆的切线有几条；

思路，求出两个圆心之间的距离，比较5种情况：重合输出-1  无数条， 相离 4 ，外切 3 ， 相交 2， 内切  1， 内含 0.

①两圆外离 d＞R+r 
②两圆外切 d=R+r 
③两圆相交 R-r＜d＜R+r(R＞r) 
④两圆内切 d=R-r(R＞r) 
⑤两圆内含d＜R-r(R＞r)

 

#include <iostream>

using namespace std;

int T,ans;
double r1,r2,x1,x2,y1,y2;
double a1,a2,b1,b2;
int main()
{
    int ans;
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--)
    {
        cin>>x1>>y1>>r1>>x2>>y2>>r2;
        double d=(x2-x1)*(x2-x1)+(y2-y1)*(y2-y1);

        if(d>(r1+r2)*(r1+r2)) ans=4;
        else if(d==(r1+r2)*(r1+r2)) ans=3;
        else if(d<(r1+r2)*(r1+r2)&&d>(r1-r2)*(r1-r2)) ans=2;
        else if(d==(r1-r2)*(r1-r2)&&(x1!=x2||y1!=y2||r1!=r2)) ans=1;
        else if(d<(r1-r2)*(r1-r2)&&(x1!=x2||y1!=y2||r1!=r2)) ans=0;
        else ans=-1;

        cout<<ans<<endl;
    }
    return 0;
}