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This is a well-known and very old fact: https://en.wikipedia.org/wiki/Tusi_couple
This problem is kinematic. Let's solve it by kinematic means.
At the picture ##O## is a center of the big circle; ##C## is a contact point of the circles. We must prove that a point ##A## on the rim of the small circle runs along a straight line while the small circle rolls inside the big circle without slipping.
It is sufficient to check that the velocity ##\boldsymbol v_A## of the point ##A## is parallel to the vector ##\boldsymbol{OA}## for all the time. The non-slippery condition implies ##\boldsymbol v_A=\boldsymbol\omega\times\boldsymbol{CA}.##
(##\boldsymbol\omega## is an angular velocity of the small circle. This vector is perpendicular to the plane of the picture)
Thus we have
$$\boldsymbol{OA}\times\boldsymbol v_A=\boldsymbol{OA}\times(\boldsymbol\omega\times \boldsymbol{CA})$$
$$=\boldsymbol\omega(\boldsymbol{OA,CA})-\boldsymbol{CA}(\boldsymbol{OA},\boldsymbol\omega)=0.$$
Here we use the BAC-CAB rule and the well-known property of a triangle inscribed in the circle.
This problem is kinematic. Let's solve it by kinematic means.
At the picture ##O## is a center of the big circle; ##C## is a contact point of the circles. We must prove that a point ##A## on the rim of the small circle runs along a straight line while the small circle rolls inside the big circle without slipping.
It is sufficient to check that the velocity ##\boldsymbol v_A## of the point ##A## is parallel to the vector ##\boldsymbol{OA}## for all the time. The non-slippery condition implies ##\boldsymbol v_A=\boldsymbol\omega\times\boldsymbol{CA}.##
(##\boldsymbol\omega## is an angular velocity of the small circle. This vector is perpendicular to the plane of the picture)
Thus we have
$$\boldsymbol{OA}\times\boldsymbol v_A=\boldsymbol{OA}\times(\boldsymbol\omega\times \boldsymbol{CA})$$
$$=\boldsymbol\omega(\boldsymbol{OA,CA})-\boldsymbol{CA}(\boldsymbol{OA},\boldsymbol\omega)=0.$$
Here we use the BAC-CAB rule and the well-known property of a triangle inscribed in the circle.
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