解二元二次方程组问题，急！！

Question

问题：已知大圆R与两小圆R1，R2内切，且两小圆R1，R2外切。
并已知大圆R的半径r0、圆心坐标x0,y0，小圆R1的半径r1、圆心坐标x1,y1，以及小圆R2半径r2。求R2圆心坐标。即已知r0,x0，y0,r1,x1,y1,r2,
列方程组
(x0-x2)*(x0-x2)+(y0-y2)*(y0-y2)=(r0-r2)*(r0-r2)
(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)=(r1+r2)*(r1+r2)求x2,y2.求代码...

Answer 1

晕 方程组都出来了，你自己写个函数
(x0-x2)*(x0-x2)+(y0-y2)*(y0-y2)=(r0-r2)*(r0-r2)
笔算，怎么算的，你函数就怎么写，很简单的啊

Answer 2

用matlab(好像是的,现在全忘了)
直接在里面画图,结果就出来了

Answer 3

我用牛顿松弛法...异常。
可能是语法上的错误，偶是java新手...
望高手解惑。
import static java.lang.Math.*;public class Account {
    private double x2=0;
    private double y2=0;
    
    public static void main(String[] args) {
        double x0 = 512;
        double y0 = 384;
          
        double x1 = 512;
        double y1 = 284;
        
        double r0 = 300;
        double r1 = 100;
        double r2 = 50;
        
        Account Third = new Account(x0,y0,x1,y1,r0,r1,r2);
        
        System.out.println(Third.getX()+","+Third.getY());
    }
    
    public Account(double x0,double y0,double x1,double y1,double r0,double r1,double r2){
        
        double[] x=new double[10000];
        double[] y=new double[10000];
        double[] j=new double[10000];
        double resultX=0;
        double resultY=0;
        
        x[0]=1;
        y[0]=1;        for(int i=0;i<j.length;i++){
            j[i]=4*((x[i]-x0)*(y[i]-y1)-(x[i]-x1)*(y[i]-y0));
        
            if(j[i]!=0) {
                x[i+1]=x[i]+2*((y[i]-y0)*f2(x[i],y[i],x1,y1,r0,r1)-(y[i]-y1)*f1(x[i],y[i],x0,y0,r0,r2))/j[i];
                y[i+1]=y[i]+2*((x[i]-x1)*f1(x[i],y[i],x0,y0,r0,r2)-(x[i]-x0)*f2(x[i],y[i],x1,y1,r0,r1))/j[i];
                
                resultX=abs(x[i+1]-x[i]);
                resultY=abs(y[i+1]-y[i]);
            }
            else {
                System.err.print("错误!");
            }
            
            if(max(resultX,resultY)<0.1){
                double x2=x[i+1];
                double y2=y[i+1];
                break;
            }
        }
    }
    
    public double getX(){
        return x2;
    }
    public double getY(){
        return y2;
    }
            
    double f1(double x,double y,double x0,double y0,double r0,double r2){
        double result1=(x-x0)*(x-x0)+(y-y0)*(y-y0)-(r0-r2)*(r0-r2);
        return result1;
    }
        
    double f2(double x,double y,double x1,double y1,double r0,double r1){
        double result2=(x-x1)*(x-x1)+(y-y1)*(y-y1)-(r0+r1)*(r0+r1);
        return result2;
    }
}