Java 从三个点找到圆心的算法是什么？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/4103405/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
What is the algorithm for finding the center of a circle from three points?
提问by Russell Strauss
I have three points on the circumference of a circle:
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);
How do I calculate the center of the circle?
如何计算圆心？
Implementing it in Processing (Java).
在处理（Java）中实现它。
I found the answer and implemented a working solution:
我找到了答案并实施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;
}
采纳答案by Azmisov
It can be a rather in depth calculation. There is a simple step-by-step here: http://paulbourke.net/geometry/circlesphere/. Once you have the equation of the circle, you can simply put it in a form involving H and K. The point (h,k) will be the center.
它可以是一个相当深入的计算。这里有一个简单的分步说明：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) 将是中心。
(scroll down a little ways at the link to get to the equations)
（在链接处向下滚动一点以获得方程式）
回答by andersoj
Here's my Java port, dodging the error condition when the determinant disappears with a very elegant IllegalArgumentException, my approach to coping with the "points are two far apart" or "points lie on a line" conditions. Also, this computes the radius (and copes with exceptional conditions) which your intersecting-slopes approach will not do.
这是我的 Java 端口，当行列式消失时IllegalArgumentException，我用一种非常优雅的. 此外，这会计算您的相交坡度方法无法执行的半径（并应对特殊情况）。
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);
}
}
See 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));
}
回答by IamRKhanna
I was looking for a similar algorithm when I hovered over this question. Took your code but found that this will not work in cases when where either of the slope is 0 or infinity (can be true when xDelta_a or xDelta_b is 0).
当我将鼠标悬停在这个问题上时，我正在寻找类似的算法。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 时为真）。
I corrected the algorithm and here is my code. Note: I used objective-c programming language and am just changing the code for point value initialization, so if that is wrong, I am sure programmer working in java can correct it. The logic, however, is the same for all (God bless algorithms!! :))
我更正了算法，这是我的代码。注意：我使用了objective-c 编程语言并且只是更改了点值初始化的代码，所以如果这是错误的，我相信使用java 工作的程序员可以更正它。然而，逻辑对所有人都是一样的（上帝保佑算法！！:)）
Works perfectly fine as far as my own functional testing is concerned. Please let me know if logic is wrong at any point.
就我自己的功能测试而言，效果很好。请让我知道是否有任何逻辑错误。
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;
}
回答by Ernesto Alfonso
public Vector2 CarculateCircleCenter(Vector2 p1, Vector2 p2, Vector2 p3)
{
if (
p2.x - p1.x == 0 ||
p3.x - p2.x == 0 ||
p2.y - p1.y == 0 ||
p3.y - p2.y == 0
) return null;
Vector2 center = new Vector2();
float ma = (p2.y - p1.y) / (p2.x - p1.x);
float mb = (p3.y - p2.y) / (p3.x - p2.x);
center.x = (ma * mb * (p1.y - p3.y) + mb * (p1.x - p2.x) - ma * (p2.x + p3.x)) / (2 * (mb - ma));
center.y = (-1 / ma) * (center.x - (p1.x + p2.x) * 0.5) + (p1.y + p2.y) * 0.5;
return center;
}
回答by Adam Gawne-Cain
I am sorry my answer is late. Any solution using "slope" will fail when two of the points form a vertical line, because the slope will be infinite.
很抱歉我的回答晚了。当两个点形成一条垂直线时，任何使用“斜率”的解决方案都将失败，因为斜率将是无限的。
Here is a simple robust solution for 2019 that always works correctly:
这是 2019 年的一个简单而强大的解决方案，它始终可以正常工作：
public static boolean circleCenter(double[] p1, double[] p2, double[] p3, double[] center) {
double ax = (p1[0] + p2[0]) / 2;
double ay = (p1[1] + p2[1]) / 2;
double ux = (p1[1] - p2[1]);
double uy = (p2[0] - p1[0]);
double bx = (p2[0] + p3[0]) / 2;
double by = (p2[1] + p3[1]) / 2;
double vx = (p2[1] - p3[1]);
double vy = (p3[0] - p2[0]);
double dx = ax - bx;
double dy = ay - by;
double vu = vx * uy - vy * ux;
if (vu == 0)
return false; // Points are collinear, so no unique solution
double g = (dx * uy - dy * ux) / vu;
center[0] = bx + g * vx;
center[1] = by + g * vy;
return true;
}
The above code will return "false" if and only if the 3 points are collinear.
当且仅当 3 个点共线时，上面的代码将返回“false”。
回答by Nitin Jindal
def circle_mid_point(list_b):
list_k = []
for i in range(3):
for j in range(1,4):
if i+j <=3:
midpoint_x1 = (list_b[i][0] + list_b[i+j][0])/2
midpoint_y1 = (list_b[i][1] + list_b[i + j][1]) / 2
list_k.append([midpoint_x1,midpoint_y1]) # list of all the midpoints of the lines
for k in range(len(list_k)):
for j in range(1,len(list_k)-1):
if list_k[k] == list_k[k+j]: #at centre only two midpoints will have the same value
return list_k[k]
k = circle_mid_point([[0,1],[1,0],[-1,0],[0,-1]])
print(k)
