如何在圆的周长上画圆？

Question

我正在尝试绘制这样的图：我不知道如何在 for 循环中找到较小圆圈的中心。首先，我尝试用较少数量的圆圈（例如2个）来绘制它，但我不知道为什么较小的圆圈是半圆？我的尝试：import numpy as npimport matplotlib.pyplot as pltr = 2, h = 1, k = 1axlim = r + np.max((abs(h),np.max(abs(k))))x = np.linspace(-axlim, axlim, 100)X,Y = np.meshgrid(x,x)F = (X-h)**2 + (Y-k)**2 - r**2plt.contour(X,Y,F,0)F1 = (X-(h+r))**2 + (Y-k)**2 - (r/3)**2plt.contour(X,Y,F1,0)F2 = (X-h)**2 + (Y-(k+r))**2 - (r/3)**2plt.contour(X,Y,F2,0)plt.gca().set_aspect('equal')plt.axis([-4*r, 4*r, -4*r,4*r])# plt.axis('off')plt.show()输出：

Answer 1

0, 2pi可以使用正弦、余弦和在范围内均匀划分的角度：

import numpy as np

import matplotlib.pyplot as plt

num_circ = 7

rad_large = 7

rad_small = 6

thetas = np.linspace(0, 2 * np.pi, num_circ, endpoint=False)

fig, ax = plt.subplots()

ax.add_patch(plt.Circle((0, 0), rad_large, fc='none', ec='navy'))

for theta in thetas:

    ax.add_patch(plt.Circle((rad_large * np.cos(theta), rad_large * np.sin(theta),), rad_small, fc='none', ec='crimson'))

ax.autoscale_view() # calculate the limits for the x and y axis

ax.set_aspect('equal') # show circles as circles

plt.show()