## Wednesday, 9 May 2012

### Uniformly Distributed Random Points Inside a Circle

So I just started my blog and I already posted my first blog. Now I want to get my hand dirty by posting some math!

The problem I want to consider here is generation of uniformly distributed random points inside different geometric shapes. The simplest shape to consider is a rectangle. In this case, the generation of uniformly distributed random points is trivial. Let $$l$$ be the length of the square, $$w$$ be the width, and the point $$p_c(x_c,y_c)$$ be the center of the rectangle. Let $$p_i(X,Y)$$ be a random point inside the rectangle. Here $$X$$ and $$Y$$ are random variables. For $$p_i$$ to be uniformly distributed, $$X$$ must be a uniform distribution over the interval $$\left[ l-\frac{x_c}{2}, l+\frac{x_c}{2} \right]$$, and $$Y$$ must be a uniform distribution over the interval $$\left[ w-\frac{y_c}{2}, w+\frac{y_c}{2} \right]$$. Well this is fairly trivial.

Now lets consider a circle. The problem is not as trivial. We define a circle by its radius $$r_c$$ and the point representing its center $$p_c(x_c,y_c)$$. For simplicity lets assume the center of the circle is on the origin (i.e. $$p_c(0, 0)$$ ). The first guess that people typically give (which is wrong) is to use polar coordinates with $$r$$ and $$\theta$$ uniformly distributed over $$\left[ 0, r_c \right]$$ and $$\left[ 0, 2\pi \right]$$, respectively. The reason this turn out to be wrong is because the area of a circle with radius $$\frac{r_c}{2}$$ and the area of a ring (or doughnut) with width $$\frac{r_c}{2}$$ are not the same. What is the correct answer? (I will post the answer with derivation in a few days!)