找到两个圆的交点

Question

假设所有其他值都已知，我怎样才能编写一个解决x ₃ 和y ₃ 的程序用于以下等式？我将用Javascript编写这个，但任何类型的伪代码都可以。应该有1,2或0个解决方案。 （正如你所看到的，我试图找出代码来找到两个圆圈的交集。）

（x ₃ - x ₁ ） ² +（y ₃ - y _{1 < / sub>） ² = r 1 ²}

和

（x ₃ - x ₂ ） ² +（y ₃ - y _{2 < / sub>） ² = r 2 ²}

Answer 1

我找到了自己问题的解决方案：

/* circle_circle_intersection() *
 * Determine the points where 2 circles in a common plane intersect.
 *
 * int circle_circle_intersection(
 *                                // center and radius of 1st circle
 *                                double x0, double y0, double r0,
 *                                // center and radius of 2nd circle
 *                                double x1, double y1, double r1,
 *                                // 1st intersection point
 *                                double *xi, double *yi,              
 *                                // 2nd intersection point
 *                                double *xi_prime, double *yi_prime)
 *
 * This is a public domain work. 3/26/2005 Tim Voght
 *
 */
#include <math.h>

int circle_circle_intersection(double x0, double y0, double r0,
                               double x1, double y1, double r1,
                               double *xi, double *yi,
                               double *xi_prime, double *yi_prime)
{
    double a, dx, dy, d, h, rx, ry;
    double x2, y2;

    /* dx and dy are the vertical and horizontal distances between
     * the circle centers.
     */
    dx = x1 - x0;
    dy = y1 - y0;

    /* Determine the straight-line distance between the centers. */
    //d = sqrt((dy*dy) + (dx*dx));
    d = hypot(dx,dy); // Suggested by Keith Briggs

    /* Check for solvability. */
    if (d > (r0 + r1))
    {
        /* no solution. circles do not intersect. */
        return 0;
    }
    if (d < fabs(r0 - r1))
    {
        /* no solution. one circle is contained in the other */
        return 0;
    }

    /* 'point 2' is the point where the line through the circle
     * intersection points crosses the line between the circle
     * centers.  
     */

    /* Determine the distance from point 0 to point 2. */
    a = ((r0*r0) - (r1*r1) + (d*d)) / (2.0 * d) ;

    /* Determine the coordinates of point 2. */
    x2 = x0 + (dx * a/d);
    y2 = y0 + (dy * a/d);

    /* Determine the distance from point 2 to either of the
     * intersection points.
     */
    h = sqrt((r0*r0) - (a*a));

    /* Now determine the offsets of the intersection points from
     * point 2.
     */
    rx = -dy * (h/d);
    ry = dx * (h/d);

    /* Determine the absolute intersection points. */
    *xi = x2 + rx;
    *xi_prime = x2 - rx;
    *yi = y2 + ry;
    *yi_prime = y2 - ry;

    return 1;
}