Solution

Let the circles initially touch at point $A$, lying on the $Ox$ axis, where point $O$ is the center of the large circle. The coordinates of point $A$ are $(kr, 0)$, where $k = \frac{R}{r}$. Let us consider how the coordinates of point $A$, tied to the rolling circle ($A$ goes to $A'$), change. Let the small circle roll so that its center moves from point $C'$ to point $C''$ and rotates relative to point $O$ by angle $t$.

First, it can be shown that the rotation of the small circle relative to its center (i.e. the angle between $CA$ and $C' A'$) is equal to $t - kt = -(k-1)t$. Second, the coordinates of point $C'$ will be as follows: $(cos(t)(k-1)r, sin(t)(k-1)r)$. Then, knowing where the center of the rolling circle will go and at what angle it has turned relative to this center, we can write down the coordinates of point $A'$:

$ \begin{cases} x = cos(t)(k-1)r + cos((k-1)t)r \\ y = sin(t)(k-1)r - sin((k-1)t)r \end{cases} $

    <h4>Answer</h4>
    <p>
        The trajectory of the wheel rim point passes along the cylinder diameter.
    </p>


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>