Consider the following diagram:

As we can see, we may state:

\(\displaystyle \tan(\theta)=\frac{x}{y}\)

Now, let's differentiate with respect to time \(t[\), bearing in mind that while \(x\) is a function of \(t\), \(y\) is a constant.

\(\displaystyle \sec^2(\theta)\cdot\d{\theta}{t}=\frac{1}{y} \cdot\d{x}{t}\)

Since we are being asked to find the speed of the spot, whose position is \(x\), we want to solve for \(\displaystyle \d{x}{t}\):

\(\displaystyle \d{x}{t}=y\sec^2(\theta) \cdot\d{\theta}{t}\)

Let's let the angular velocity be given by:

\(\displaystyle \omega=\d{\theta}{t}\)

and from the diagram and the Pythagorean theorem, we find:

\(\displaystyle \sec^2(\theta)=\frac{x^2+y^2}{y^2}\)

Hence, we have:

\(\displaystyle \d{x}{t}=\frac{\omega}{y}(y^2+x^2)\)

Now, the angular velocity is in radians per unit of time, and since there are \(2\pi\) radians per revolution, we may write:

\(\displaystyle \omega=2\pi R\)

Hence:

\(\displaystyle \d{x}{t}=\frac{2\pi R}{y}(y^2+x^2)\)