I think the "trick" to this is that the "inner circle" (polygon) has exactly half the radius of the outer circle.
Imagine we trace the path of a point on a circle of radius $r$ as it travels inside a circle of radius $2r$. Since it doesn't really matter "when" we start tracking it (the path is periodic), assume that both circles are touching at the point $(0,2r)$ at $t = 0$, and that the outer circle is centered at the origin.
As the inner circle "rolls" counter-clockwise, the point on the inner circle we are tracking moves CLOCKWISE around a shifting center.
This center is at: $((2r-r)\cos t,(2r-r)\sin t) = (r\cos t,r\sin t)$. Since the outer circle's circumference (which is directly proportional to radius) is twice that of the inner circle, as the center has moved through an angle of $t$, the point we are tracking makes an angle of $2t$ with the point of tangency. Half of this angle is $t$, the other half is the angle our tracked point makes to a horizontal line passing through the center of the inner circle.
It follows our tracked point has coordinates:
$(r\cos(-t),r\sin(-t)) + (r\cos t,r\sin t) = (2r\cos t,0)$.
As $t$ varies, the image $\{(x(t),y(t)): t \in \Bbb R_0^+\}$ is the interval $[-2r,2r]\times \{0\}$, which is a "straight-line" (segment).
