various constructions of the Limaçon curve

limacon-1.md

Wikipedia: Let P be a point and C be a circle whose center is not P. Then the envelope of those circles whose center lies on C and that pass through P is a limaçon.

limacon-1.py

import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
plt.rcParams['figure.figsize']=6,6

def myCircle(z,r, N=50):
    dt = 1.0/N
    t = np.arange(0,1+dt,dt)
    return z + r*np.exp(2j*np.pi*t)

N = 50
W = myCircle(0,1)
plt.plot(W.real, W.imag, linewidth=2, c='#C96640')

plt.plot([0],[0], 'o', markersize=5)

P = 2j*np.exp(2j*np.pi*0)

plt.plot(P.real,P.imag, 'o',markersize=5)

for w in W[:]:
    b = myCircle( 0.5*(w + P), np.abs(0.5*(w-P)))
    plt.plot(b.real, b.imag, '#798B6E', alpha=0.35)

plt.axis("Equal")
plt.grid(True)

limacon-2.md

limacon-2.py

import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
plt.rcParams['figure.figsize']=6,6

N = 400
w = myCircle(0,1,N=N)
P = 2j


pts = []

for k in range(N):

    a = w[k] + 2j*w[k]
    b = w[k] - 2j*w[k]
    Q = P - (((P)*np.conj(w[k])).real - 1)*w[k]
    plt.plot([P.real, Q.real], [P.imag, Q.imag], 'r-',markersize=5, alpha=0.5)
    
    pts += [Q]
    
pts = np.array(pts)
plt.plot(pts.real, pts.imag, '-')


plt.axis("Equal")
plt.grid(True)

plt.xlim([-2,2])
plt.ylim([-1.5,2.5])