Students will learn the sector and arc length formulas for circles, and practice using them. Then give your students this challenge. And after that, this one:

Problem: There is a small circle with center \(Y\) and radius \(r\) rolling around inside a large circle with center \(X\) and radius \(2r.\) Let \(C\) be the point where the two circles are in contact. A pen is fixed somewhere on the circumference of the smaller circle. Let's call the position of the pen \(P.\) Let \(D\) be the point where the ray \(XP\) intersects the large circle. Prove that arc \(CP\) has the same length as \(CD.\) This will show that P always lies on the diameter \(ED.\)

Solution: First, notice that \(\overline{XY} \cong \overline{PY}\) because both are radii of the smaller circle. That makes \(XYP\) an isosceles triangle. Let's call its base angles \(\theta.\) Hence, by the exterior angle theorem, \(\angle PYC = 2\theta.\) Using the arc length formula, we find

$$\begin{align} CP = (2r)(\theta) = 2r\theta \\ CD = (r)(2\theta) = 2r\theta \end{align}$$Here's the source of the problem and solution. I found both lacking, which is why I've altered them slightly.