In python how to generate a random pair of points (x,y) that lies inside a circle of radius r.
Basically the x and y should satisfy the condition x^2 + y^2 = r^2.
2 Answers
To generate uniformly distributed point inside origin-centered circle of radius r, you can generate two uniform values t,u in range 0..1 and use the next formula:
import math, random
r = 4
t = random.random()
u = random.random()
x = r * math.sqrt(t) * math.cos(2 * math.pi * u)
y = r * math.sqrt(t) * math.sin(2 * math.pi * u)
print (x,y)
3 Comments
Kaustubh Dwivedi
What changes can be made such that x and y lie between the region of two concentric circles of radius r1 and r2. r1^2 <= x^2 + y^2 <= r2^2
FBruzzesi
@MBo I think
t should be in range Uniform[r1**2/r2**2, 1]
MBo
For ring - generate value t in range (r1/r2)**2..1 with RandomUniform and use x = r2 * math.sqrt(t)... Corresponding answer
Using numpy to generate more than one point at a time:
import numpy as np
import matplotlib.pyplot as plt
n_samples = 1000
r = 4
# make a simple unit circle
theta = np.linspace(0, 2*np.pi, n_samples)
a, b = r * np.cos(theta), r * np.sin(theta)
t = np.random.uniform(0, 1, size=n_samples)
u = np.random.uniform(0, 1, size=n_samples)
x = r*np.sqrt(t) * np.cos(2*np.pi*u)
y = r*np.sqrt(t) * np.sin(2*np.pi*u)
# Plotting
plt.figure(figsize=(7,7))
plt.plot(a, b, linestyle='-', linewidth=2, label='Circle', color='red')
plt.scatter(x, y, marker='o', label='Samples')
plt.ylim([-r*1.5,r*1.5])
plt.xlim([-r*1.5,r*1.5])
plt.grid()
plt.legend(loc='upper right')
plt.show(block=True)
which results in:
inside a circle of radius rassumes x^2 + y^2 <= r^2.