Generate a random point within a circle (uniformly)

I need to generate a uniformly random point within a circle of radius R.

I realize that by just picking a uniformly random angle in the interval [0 ... 2π), and uniformly random radius in the interval (0 ... R) I would end up with more points towards the center, since for two given radii, the points in the smaller radius will be closer to each other than for the points in the larger radius.

I found a blog entry on this over here but I don't understand his reasoning. I suppose it is correct, but I would really like to understand from where he gets (2/R²)×r and how he derives the final solution.

Update: 7 years after posting this question I still hadn't received a satisfactory answer on the actual question regarding the math behind the square root algorithm. So I spent a day writing an answer myself. Link to my answer.

How to generate a random point within a circle of radius R:

r = R * sqrt(random())
theta = random() * 2 * PI

(Assuming random() gives a value between 0 and 1 uniformly)

If you want to convert this to Cartesian coordinates, you can do

x = centerX + r * cos(theta)
y = centerY + r * sin(theta)

Why sqrt(random())?

Let's look at the math that leads up to sqrt(random()). Assume for simplicity that we're working with the unit circle, i.e. R = 1.

The average distance between points should be the same regardless of how far from the center we look. This means for example, that looking on the perimeter of a circle with circumference 2 we should find twice as many points as the number of points on the perimeter of a circle with circumference 1.

Since the circumference of a circle (2πr) grows linearly with r, it follows that the number of random points should grow linearly with r. In other words, the desired probability density function (PDF) grows linearly. Since a PDF should have an area equal to 1 and the maximum radius is 1, we have

So we know how the desired density of our random values should look like. Now: How do we generate such a random value when all we have is a uniform random value between 0 and 1?

We use a trick called inverse transform sampling

1. From the PDF, create the cumulative distribution function (CDF)
2. Mirror this along y = x
3. Apply the resulting function to a uniform value between 0 and 1.

Sounds complicated? Let me insert a blockquote with a little side track that conveys the intuition:

Suppose we want to generate a random point with the following distribution:

                

That is

• 1/5 of the points uniformly between 1 and 2, and
• 4/5 of the points uniformly between 2 and 3.

The CDF is, as the name suggests, the cumulative version of the PDF. Intuitively: While PDF(x) describes the number of random values at x, CDF(x) describes the number of random values less than x.

In this case the CDF would look like:

                

To see how this is useful, imagine that we shoot bullets from left to right at uniformly distributed heights. As the bullets hit the line, they drop down to the ground:

                

See how the density of the bullets on the ground correspond to our desired distribution! We're almost there!

The problem is that for this function, the y axis is the output and the x axis is the input. We can only "shoot bullets from the ground straight up"! We need the inverse function!

This is why we mirror the whole thing; x becomes y and y becomes x:

                

We call this CDF^-1. To get values according to the desired distribution, we use CDF^-1(random()).

…so, back to generating random radius values where our PDF equals 2x.

Step 1: Create the CDF:

Since we're working with reals, the CDF is expressed as the integral of the PDF.

CDF(x) = ∫ 2x = x²

Step 2: Mirror the CDF along y = x:

Mathematically this boils down to swapping x and y and solving for y:

CDF:     y = x²
Swap:   x = y²
Solve:   y = √x
CDF^-1:  y = √x

Step 3: Apply the resulting function to a uniform value between 0 and 1

CDF^-1(random()) = √random()

Which is what we set out to derive :-)

Let's approach this like Archimedes would have.

How can we generate a point uniformly in a triangle ABC, where |AB|=|BC|? Let's make this easier by extending to a parallelogram ABCD. It's easy to generate points uniformly in ABCD. We uniformly pick a random point X on AB and Y on BC and choose Z such that XBYZ is a parallelogram. To get a uniformly chosen point in the original triangle we just fold any points that appear in ADC back down to ABC along AC.

Now consider a circle. In the limit we can think of it as infinitely many isoceles triangles ABC with B at the origin and A and C on the circumference vanishingly close to each other. We can pick one of these triangles simply by picking an angle theta. So we now need to generate a distance from the center by picking a point in the sliver ABC. Again, extend to ABCD, where D is now twice the radius from the circle center.

Picking a random point in ABCD is easy using the above method. Pick a random point on AB. Uniformly pick a random point on BC. Ie. pick a pair of random numbers x and y uniformly on [0,R] giving distances from the center. Our triangle is a thin sliver so AB and BC are essentially parallel. So the point Z is simply a distance x+y from the origin. If x+y>R we fold back down.

Here's the complete algorithm for R=1. I hope you agree it's pretty simple. It uses trig, but you can give a guarantee on how long it'll take, and how many random() calls it needs, unlike rejection sampling.

t = 2*pi*random()
u = random()+random()
r = if u>1 then 2-u else u
[r*cos(t), r*sin(t)]

Here it is in Mathematica.

f[] := Block[{u, t, r},
  u = Random[] + Random[];
  t = Random[] 2 Pi;
  r = If[u > 1, 2 - u, u];
  {r Cos[t], r Sin[t]}
]

ListPlot[Table[f[], {10000}], AspectRatio -> Automatic]

Here is a fast and simple solution.

Pick two random numbers in the range (0, 1), namely a and b. If b < a, swap them. Your point is (b*R*cos(2*pi*a/b), b*R*sin(2*pi*a/b)).

You can think about this solution as follows. If you took the circle, cut it, then straightened it out, you'd get a right-angled triangle. Scale that triangle down, and you'd have a triangle from (0, 0) to (1, 0) to (1, 1) and back again to (0, 0). All of these transformations change the density uniformly. What you've done is uniformly picked a random point in the triangle and reversed the process to get a point in the circle.