java中已知两圆坐标及半径，求两圆外切线的坐标

Question

已知两圆坐标及半径，想求两个圆的外切线的四个坐标，不知道用java代码该怎么实现了。下面是我的实现，出来的结果一直不对：

 public static List<Point> getPointLocations(Point p1, Point p2){
        List<Point> pointList = new LinkedList<>();
        double k = (double)(p2.y-p1.y)/(double) (p2.x-p1.x);//求出斜率
        double c = p1.y-k*p1.x;//求出p1点的常数系数
        double k2 = -(1/k);//垂直于这条线的斜率
        double c2 = p1.y+p1.x/k;//垂直线的斜率常数系数
        int size = 45;//半径
        double c3 = size * (Math.sqrt(k*k+1))+c;
        double c4 = c - size * Math.sqrt(k*k+1);

        Log.e(TAG, "getPointLocations: k:"+k+"  c:"+c+"   k2:"+k2+"  c2:"+c2+"   c3:"+c3+"  c4:"+c4);

        double x1 = (c2 - c3)/(k+k2);
        double y1 = k*x1+c3;

        double x2 = (c2 - c4)/(k+k2);
        double y2 = k*x2+c4;

        double p2c = p2.y-k*p2.x;
        double p2c2 = p2.y+p2.x/k;
        double p2c3 = size * Math.sqrt(k*k+1)+p2c;
        double p2c4 = p2c - size * (Math.sqrt(k*k+1));

        double x3 = (p2c2 - p2c3)/(k+k2);
        double y3 = k*x3+p2c3;

        double x4 = (p2c2 - p2c4)/(k+k2);
        double y4 = k*x4+p2c4;


        pointList.add(new Point((int)(x1-p1.x),(int)(y1-p1.y)));
        pointList.add(new Point((int)(x2-p1.x),(int)(y2-p1.y)));
        pointList.add(new Point((int)(x3-p2.x),(int)(y3-p2.y)));
        pointList.add(new Point((int)(x4-p2.x),(int)(y4-p2.y)));

        Log.e("TAG", "getPointLocations: "+"  size   "+size+"   "+pointList.toString() );

        return pointList;
    }

Answer 1

没理解你的意思，怎么会有四个点呢

Answer 2

参考：https://blog.csdn.net/victor_woo/article/details/10065585