我曾经用过这个方法: 这可能是完全未优化的(即它使用了一个点数组，所以它不能用于大圆圈)，但它提供了足够的随机分布。如果你愿意，你可以跳过矩阵的创建，直接绘制。方法是随机化矩形中落在圆内的所有点。

bool[,] getMatrix(System.Drawing.Rectangle r) {
    bool[,] matrix = new bool[r.Width, r.Height];
    return matrix;
}

void fillMatrix(ref bool[,] matrix, Vector center) {
    double radius = center.X;
    Random r = new Random();
    for (int y = 0; y < matrix.GetLength(0); y++) {
        for (int x = 0; x < matrix.GetLength(1); x++)
        {
            double distance = (center - new Vector(x, y)).Length;
            if (distance < radius) {
                matrix[x, y] = r.NextDouble() > 0.5;
            }
        }
    }

}

private void drawMatrix(Vector centerPoint, double radius, bool[,] matrix) {
    var g = this.CreateGraphics();

    Bitmap pixel = new Bitmap(1,1);
    pixel.SetPixel(0, 0, Color.Black);

    for (int y = 0; y < matrix.GetLength(0); y++)
    {
        for (int x = 0; x < matrix.GetLength(1); x++)
        {
            if (matrix[x, y]) {
                g.DrawImage(pixel, new PointF((float)(centerPoint.X - radius + x), (float)(centerPoint.Y - radius + y)));
            }
        }
    }

    g.Dispose();
}

private void button1_Click(object sender, EventArgs e)
{
    System.Drawing.Rectangle r = new System.Drawing.Rectangle(100,100,200,200);
    double radius = r.Width / 2;
    Vector center = new Vector(r.Left + radius, r.Top + radius);
    Vector normalizedCenter = new Vector(radius, radius);
    bool[,] matrix = getMatrix(r);
    fillMatrix(ref matrix, normalizedCenter);
    drawMatrix(center, radius, matrix);
}

我曾经用过这个方法: 这可能是完全未优化的(即它使用了一个点数组，所以它不能用于大圆圈)，但它提供了足够的随机分布。如果你愿意，你可以跳过矩阵的创建，直接绘制。方法是随机化矩形中落在圆内的所有点。

bool[,] getMatrix(System.Drawing.Rectangle r) {
    bool[,] matrix = new bool[r.Width, r.Height];
    return matrix;
}

void fillMatrix(ref bool[,] matrix, Vector center) {
    double radius = center.X;
    Random r = new Random();
    for (int y = 0; y < matrix.GetLength(0); y++) {
        for (int x = 0; x < matrix.GetLength(1); x++)
        {
            double distance = (center - new Vector(x, y)).Length;
            if (distance < radius) {
                matrix[x, y] = r.NextDouble() > 0.5;
            }
        }
    }

}

private void drawMatrix(Vector centerPoint, double radius, bool[,] matrix) {
    var g = this.CreateGraphics();

    Bitmap pixel = new Bitmap(1,1);
    pixel.SetPixel(0, 0, Color.Black);

    for (int y = 0; y < matrix.GetLength(0); y++)
    {
        for (int x = 0; x < matrix.GetLength(1); x++)
        {
            if (matrix[x, y]) {
                g.DrawImage(pixel, new PointF((float)(centerPoint.X - radius + x), (float)(centerPoint.Y - radius + y)));
            }
        }
    }

    g.Dispose();
}

private void button1_Click(object sender, EventArgs e)
{
    System.Drawing.Rectangle r = new System.Drawing.Rectangle(100,100,200,200);
    double radius = r.Width / 2;
    Vector center = new Vector(r.Left + radius, r.Top + radius);
    Vector normalizedCenter = new Vector(radius, radius);
    bool[,] matrix = getMatrix(r);
    fillMatrix(ref matrix, normalizedCenter);
    drawMatrix(center, radius, matrix);
}

我认为在这种情况下，使用极坐标是一种使问题复杂化的方法，如果你在一个边长为2R的正方形中随机选择点，然后选择点(x,y)使x^2+y^2<=R^2，这将会容易得多。

圆中的面积元是dA=rdr*dphi。这个额外的因子r破坏了你随机选择r和的想法。虽然phi分布平坦，但r不是，而是在1/r内平坦(也就是说，你更有可能击中边界而不是“靶心”)。

为了生成在圆上均匀分布的点从平面分布中选取r从1/r分布中选取。

或者使用Mehrdad提出的蒙特卡罗方法。

EDIT

要在1/r中选择一个随机的r，你可以从区间[1/ r，无穷]中选择一个随机的x，并计算r=1/x。R以1/ R为单位平坦分布。

为了计算一个随机的，从区间[0,1]中选择一个随机的x，并计算=2*pi*x。

朴素解不起作用的原因是它给了靠近圆中心的点更高的概率密度。换句话说，半径为r/2的圆被选中点的概率为r/2，但它的面积(点的数量)为*r^2/4。

因此，我们希望半径概率密度具有以下性质:

选择半径小于或等于给定r的概率必须与半径为r的圆的面积成正比(因为我们希望在点上有一个均匀的分布，面积越大意味着点越多)。

换句话说，我们希望在[0,r]之间选择半径的概率等于它在圆的总面积中所占的份额。圆的总面积是*R^2，半径为R的圆的面积是*R^2。因此，我们希望在[0,r]之间选择半径的概率为(pi*r^2)/(pi* r^2) = r^2/ r^2。

现在来算算:

The probability of choosing a radius between [0,r] is the integral of p(r) dr from 0 to r (that's just because we add all the probabilities of the smaller radii). Thus we want integral(p(r)dr) = r^2/R^2. We can clearly see that R^2 is a constant, so all we need to do is figure out which p(r), when integrated would give us something like r^2. The answer is clearly r * constant. integral(r * constant dr) = r^2/2 * constant. This has to be equal to r^2/R^2, therefore constant = 2/R^2. Thus you have the probability distribution p(r) = r * 2/R^2

Note: Another more intuitive way to think about the problem is to imagine that you are trying to give each circle of radius r a probability density equal to the proportion of the number of points it has on its circumference. Thus a circle which has radius r will have 2 * pi * r "points" on its circumference. The total number of points is pi * R^2. Thus you should give the circle r a probability equal to (2 * pi * r) / (pi * R^2) = 2 * r/R^2. This is much easier to understand and more intuitive, but it's not quite as mathematically sound.

这样一个有趣的问题。 一个点被选择的概率随着距离轴原点的增加而降低的基本原理在上面已经解释了多次。我们通过取U[0,1]的根来解释这一点。 下面是Python 3中正r的通解。

import numpy
import math
import matplotlib.pyplot as plt

def sq_point_in_circle(r):
    """
    Generate a random point in an r radius circle 
    centered around the start of the axis
    """

    t = 2*math.pi*numpy.random.uniform()
    R = (numpy.random.uniform(0,1) ** 0.5) * r

    return(R*math.cos(t), R*math.sin(t))

R = 200 # Radius
N = 1000 # Samples

points = numpy.array([sq_point_in_circle(R) for i in range(N)])
plt.scatter(points[:, 0], points[:,1])