从三个点找到圆的圆心的算法是什么？
从三个点找到圆的圆心的算法是什么？
提问于 2010-11-05 11:36:00
我在圆的圆周上有三个点：
pt A = (A.x, A.y);
pt B = (B.x, B.y);
pt C = (C.x, C.y);如何计算圆的圆心？
在处理中实现它(Java)。
我找到了答案并实现了一个可行的解决方案：
pt circleCenter(pt A, pt B, pt C) {
float yDelta_a = B.y - A.y;
float xDelta_a = B.x - A.x;
float yDelta_b = C.y - B.y;
float xDelta_b = C.x - B.x;
pt center = P(0,0);
float aSlope = yDelta_a/xDelta_a;
float bSlope = yDelta_b/xDelta_b;
center.x = (aSlope*bSlope*(A.y - C.y) + bSlope*(A.x + B.x)
- aSlope*(B.x+C.x) )/(2* (bSlope-aSlope) );
center.y = -1*(center.x - (A.x+B.x)/2)/aSlope + (A.y+B.y)/2;
return center;
}回答 6
Stack Overflow用户
回答已采纳
发布于 2010-11-05 11:38:59
这可能是一个相当深入的计算。这里有一个简单的步骤：http://paulbourke.net/geometry/circlesphere/。一旦你有了圆的方程，你可以简单地把它放在一个包含H和K的形式中。点(h，k)将是圆心。
(在链接中向下滚动到方程式)
Stack Overflow用户
发布于 2010-11-05 11:40:15
这是我的Java端口，用一个非常优雅的IllegalArgumentException避免了行列式消失时的错误条件，我的方法是处理“点是两个相隔很远的”或“点在一条线上”的条件。此外，这将计算半径(并处理异常情况)，这是您的相交坡度方法所不能做到的。
public class CircleThree
{
static final double TOL = 0.0000001;
public static Circle circleFromPoints(final Point p1, final Point p2, final Point p3)
{
final double offset = Math.pow(p2.x,2) + Math.pow(p2.y,2);
final double bc = ( Math.pow(p1.x,2) + Math.pow(p1.y,2) - offset )/2.0;
final double cd = (offset - Math.pow(p3.x, 2) - Math.pow(p3.y, 2))/2.0;
final double det = (p1.x - p2.x) * (p2.y - p3.y) - (p2.x - p3.x)* (p1.y - p2.y);
if (Math.abs(det) < TOL) { throw new IllegalArgumentException("Yeah, lazy."); }
final double idet = 1/det;
final double centerx = (bc * (p2.y - p3.y) - cd * (p1.y - p2.y)) * idet;
final double centery = (cd * (p1.x - p2.x) - bc * (p2.x - p3.x)) * idet;
final double radius =
Math.sqrt( Math.pow(p2.x - centerx,2) + Math.pow(p2.y-centery,2));
return new Circle(new Point(centerx,centery),radius);
}
static class Circle
{
final Point center;
final double radius;
public Circle(Point center, double radius)
{
this.center = center; this.radius = radius;
}
@Override
public String toString()
{
return new StringBuilder().append("Center= ").append(center).append(", r=").append(radius).toString();
}
}
static class Point
{
final double x,y;
public Point(double x, double y)
{
this.x = x; this.y = y;
}
@Override
public String toString()
{
return "("+x+","+y+")";
}
}
public static void main(String[] args)
{
Point p1 = new Point(0.0,1.0);
Point p2 = new Point(1.0,0.0);
Point p3 = new Point(2.0,1.0);
Circle c = circleFromPoints(p1, p2, p3);
System.out.println(c);
}
}请参阅algorithm from here
void circle_vvv(circle *c)
{
c->center.w = 1.0;
vertex *v1 = (vertex *)c->c.p1;
vertex *v2 = (vertex *)c->c.p2;
vertex *v3 = (vertex *)c->c.p3;
float bx = v1->xw; float by = v1->yw;
float cx = v2->xw; float cy = v2->yw;
float dx = v3->xw; float dy = v3->yw;
float temp = cx*cx+cy*cy;
float bc = (bx*bx + by*by - temp)/2.0;
float cd = (temp - dx*dx - dy*dy)/2.0;
float det = (bx-cx)*(cy-dy)-(cx-dx)*(by-cy);
if (fabs(det) < 1.0e-6) {
c->center.xw = c->center.yw = 1.0;
c->center.w = 0.0;
c->v1 = *v1;
c->v2 = *v2;
c->v3 = *v3;
return;
}
det = 1/det;
c->center.xw = (bc*(cy-dy)-cd*(by-cy))*det;
c->center.yw = ((bx-cx)*cd-(cx-dx)*bc)*det;
cx = c->center.xw; cy = c->center.yw;
c->radius = sqrt((cx-bx)*(cx-bx)+(cy-by)*(cy-by));
}Stack Overflow用户
发布于 2013-10-10 18:57:24
当我在这个问题上徘徊时，我正在寻找一个类似的算法。使用您的代码，但发现这在斜率为0或无穷大的情况下不起作用(当xDelta_a或xDelta_b为0时可能为真)。
我修正了算法，下面是我的代码。注意:我使用的是objective-c编程语言，并且正在修改点值初始化的代码，所以如果这是错误的，我相信在java中工作的程序员可以纠正它。然而，逻辑对所有人都是一样的(上帝保佑算法！！：)
就我自己的功能测试而言，它工作得非常好。如果逻辑有误，请随时告诉我。
pt circleCenter(pt A, pt B, pt C) {
float yDelta_a = B.y - A.y;
float xDelta_a = B.x - A.x;
float yDelta_b = C.y - B.y;
float xDelta_b = C.x - B.x;
pt center = P(0,0);
float aSlope = yDelta_a/xDelta_a;
float bSlope = yDelta_b/xDelta_b;
pt AB_Mid = P((A.x+B.x)/2, (A.y+B.y)/2);
pt BC_Mid = P((B.x+C.x)/2, (B.y+C.y)/2);
if(yDelta_a == 0) //aSlope == 0
{
center.x = AB_Mid.x;
if (xDelta_b == 0) //bSlope == INFINITY
{
center.y = BC_Mid.y;
}
else
{
center.y = BC_Mid.y + (BC_Mid.x-center.x)/bSlope;
}
}
else if (yDelta_b == 0) //bSlope == 0
{
center.x = BC_Mid.x;
if (xDelta_a == 0) //aSlope == INFINITY
{
center.y = AB_Mid.y;
}
else
{
center.y = AB_Mid.y + (AB_Mid.x-center.x)/aSlope;
}
}
else if (xDelta_a == 0) //aSlope == INFINITY
{
center.y = AB_Mid.y;
center.x = bSlope*(BC_Mid.y-center.y) + BC_Mid.x;
}
else if (xDelta_b == 0) //bSlope == INFINITY
{
center.y = BC_Mid.y;
center.x = aSlope*(AB_Mid.y-center.y) + AB_Mid.x;
}
else
{
center.x = (aSlope*bSlope*(AB_Mid.y-BC_Mid.y) - aSlope*BC_Mid.x + bSlope*AB_Mid.x)/(bSlope-aSlope);
center.y = AB_Mid.y - (center.x - AB_Mid.x)/aSlope;
}
return center;
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/4103405
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