寻找两个圆的交点 [英] Finding the intersection of two circles

问题描述

我正在尝试使用Python(使用Matplotlib)查找两个圆之间的交点，但无法获取任何值.

I'm trying to find the intersections between two circles in Python(using Matplotlib) but can't get any values back.

我这样做是通过为每个单独的圆创建X和Y的列表(Matplotlib在绘制圆时将第一个参数作为X值，第二个参数作为Y值)，然后相应地将列表相交(例如， circle1 x值和circle2 x值).

I'm doing this by creating lists of X's and Y's for each individual circle(Matplotlib takes the first argument as X values and the second one as Y values when drawing a circle), and then intersecting the lists accordingly(e.g., circle1 x values with circle2 x values).

import numpy
import math
import matplotlib.pyplot as plt
import random

def origin_circle():
    global x_points
    global y_points
    global r
    global n
    r=1
    n=2**16
    x_points=[(r*math.cos(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]
    y_points=[(r*math.sin(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]

def new_circle(x_offset, y_offset):
    global x_points1
    global y_points1
    x_points1=[x_offset+(r*math.cos(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]
    y_points1=[y_offset+(r*math.sin(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]

origin_circle()
new_center= random.randint(0, len(x_points))
x_offset = x_points[new_center]
y_offset = y_points[new_center]
new_circle(x_offset, y_offset)
print(set(x_points1).intersection(set(x_points)))
print(set(y_points1).intersection(set(y_points)))

我希望取回值，但是返回的集合为空.

I expected to get values back, but the set that returned was empty.

推荐答案

求解两个圆的交点的正确方法是代数形式.由于坐标系统(实数)的无限精度，因此无法使用点(x，y坐标)来执行此操作.

The correct method to solve for intersection points of two circles is algebraically. You can't do it using points (x, y coordinates) because of infinite precision of coordinate system (real numbers).

如果两个圆在两个点处相交，则有一种简单的方法可以计算出这两个相交点. 此处在Intersection of two circles部分下.

If two circle intersect at two points then there is straight forward way to calculate those two points of intersection. The algebra is detailed here under section Intersection of two circles.

我们还可以消除两个圆圈不相交的情况

We can also eliminate the cases when two circles are not intersecting as below

• 如果两个圆的原点之间的距离>两个圆的半径之和，则表示圆是分开的，因此不相交.
• 如果两个圆原点之间的距离<两个圆的半径之间的绝对差，则表示一个圆包含另一个圆，因此不相交.

返回两个圆的两个相交点的代码.每个小节由其中心(x，y)和半径(r)来描述

Code to return the two intersecting points of two circle. Each cricle is describe by its center (x,y) and radius (r)

def get_intercetions(x0, y0, r0, x1, y1, r1):
    # circle 1: (x0, y0), radius r0
    # circle 2: (x1, y1), radius r1

    d=math.sqrt((x1-x0)**2 + (y1-y0)**2)

    # non intersecting
    if d > r0 + r1 :
        return None
    # One circle within other
    if d < abs(r0-r1):
        return None
    # coincident circles
    if d == 0 and r0 == r1:
        return None
    else:
        a=(r0**2-r1**2+d**2)/(2*d)
        h=math.sqrt(r0**2-a**2)
        x2=x0+a*(x1-x0)/d   
        y2=y0+a*(y1-y0)/d   
        x3=x2+h*(y1-y0)/d     
        y3=y2-h*(x1-x0)/d 

        x4=x2-h*(y1-y0)/d
        y4=y2+h*(x1-x0)/d

        return (x3, y3, x4, y4)

让我们通过绘图对其进行可视化测试

# intersection circles
x0, y0 = 0, 0
r0 = 5
x1, y1 = 2, 2
r1 = 5

# intersecting with (x1, y1) but not with (x0, y0)
x2, y2 = -1,0
r2 = 2.5

circle1 = plt.Circle((x0, y0), r0, color='b', fill=False)
circle2 = plt.Circle((x1, y1), r1, color='b', fill=False)
circle3 = plt.Circle((x2, y2), r2, color='b', fill=False)

fig, ax = plt.subplots() 
ax.set_xlim((-10, 10))
ax.set_ylim((-10, 10))
ax.add_artist(circle1)
ax.add_artist(circle2)
ax.add_artist(circle3)

intersections = get_intercetions(x0, y0, r0, x1, y1, r1)
if intersections is not None:
    i_x3, i_y3, i_x4, i_y4 = intersections 
    plt.plot([i_x3, i_x4], [i_y3, i_y4], '.', color='r')

intersections = get_intercetions(x0, y0, r0, x2, y2, r2)
if intersections is not None:
    i_x3, i_y3, i_x4, i_y4 = intersections 
    plt.plot([i_x3, i_x4], [i_y3, i_y4], '.', color='r')

intersections = get_intercetions(x1, y1, r1, x2, y2, r2)
if intersections is not None:
    i_x3, i_y3, i_x4, i_y4 = intersections 
    plt.plot([i_x3, i_x4], [i_y3, i_y4], '.', color='r')

plt.gca().set_aspect('equal', adjustable='box')

输出:

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