In the diagram, a circle of radius $OA=2$ rolls around the inside of a circle of radius $OB=4$ with the half radius $AP=1$ originally positioned so that $P$ is on $OB$. Find a parameterization for the path that $P$ traces out as the inner circle rolls (without slipping.) Start by choosing an appropriate parameter. Is that a familiar curve?
The key here is to assert that the arclength swept out by rolling is the same on both circles, so if $\theta$ is the angle swept out by $\overrightarrow{OA}$ as the circle rolls and $\alpha$ is the angle swept out by $\overrightarrow{AP}$ then the arclengths are equal so $4\theta=2\alpha\Rightarrow \alpha=2\theta$. However, $\theta$ is subtracted from $\alpha$ because as the inner circle rolls, it both rotates around its center and rotates around the bigger circle's center; at the same time the inner circle's center is rotating around the bigger circle's center, so the angle from the vertical of the line $\overline{AP}$ is $2\theta-\theta=\theta$. You can think of the process of getting from the origin (0,0) to the point $P$ as a two step process: first get from (0,0) to $A=(2\sin\theta,2\cos\theta)$ and then get from $A$ to $P$, which is the vector $\overrightarrow{AP}=(\sin\theta,-\cos\theta)$, thus the coordinates of $P$ are $(3\sin\theta,\cos\theta)$. Since these coordinates satisfy the equation $\dfrac{x^2}{9}+y^2=1$ the equation represents an ellipse.