找到两个圆的交点
Find the intersection point(s) of two circles
假设所有其他值都已知,我如何编写一个程序来求解以下方程的 x3 和 y3?我打算在 Javascript 中写这个,但是任何一种伪代码都可以。应该有 1、2 或 0 个解决方案。 (如您所见,我正在尝试找出代码来找到两个圆的交点。)
(x3 - x1)2 + (y3 - y1)2 = r1 2
和
(x3 - x2)2 + (y3 - y2)2 = r2 2
我找到了我自己的问题的解决方案:
/* 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;
}
假设所有其他值都已知,我如何编写一个程序来求解以下方程的 x3 和 y3?我打算在 Javascript 中写这个,但是任何一种伪代码都可以。应该有 1、2 或 0 个解决方案。 (如您所见,我正在尝试找出代码来找到两个圆的交点。)
(x3 - x1)2 + (y3 - y1)2 = r1 2
和
(x3 - x2)2 + (y3 - y2)2 = r2 2
我找到了我自己的问题的解决方案:
/* 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;
}