Circle A is subject to both a rotation motion (about its center) and a translation motion (around the circumference of Circle B). In order to prevent slippage, the rotation motion must conveniently adjust its speed to the translation speed, depending on the relative sizes of the two circles. Now, suppose we stop the rotation and let Circle A simply glide around Circle B., glueing, as it were, the point of tangency. This means that the relative positions of the point of tangency and the center of Circle A change in such a way that, at the end of the process, they are back at its original state. Thus, Circle A will have given one whole turn, due only to the translation motion. This, I hope, explains why there is one more turn in the end. The rotation motion contributes whatever number of revolutions result from the ratio between the radii of the circles, and the translation motion contributes one additional turn.