Mathematically, let r be the radius of your circle, which defines the motion of your object. Circles are 2D figures, so we can represent the object's motion in a plane; I will use 2D Cartesian for this example. The location of its center is irrelevant (as it remains constant), so let the center be at the origin (0,0).
Therefore, the object's position can be represented via the position vector
\vec P = r\left( {\hat i\cos \theta + \hat j\sin \theta } \right)
where \hat i & \hat j represent unit vectors in the x and y directions,
respectively.
Simply put, you can represent the object position as the head of a position vector whose tail is the center of the circle, which we set at (0,0). It has a magnitude r with an "x" component equivalent to r \hat i \cos \theta and a
"y" component equivalent to r \hat j \sin \theta, where \theta is the angle swept counterclockwise from the positive x-axis, with the vertex being the origin.
Displacement is a vector quantity associated with a change in position. Where \vec P is the position vector, displacement is defined as \Delta \vec P.
In a "complete revolution", the \Delta \theta = 360^\circ = 2\pi. If we let \theta _0 represent the initial angle relevant to the object's position vector, then in a complete revolution:
\Delta \vec P = r\Delta \left( {\hat i\cos \theta + \hat j\sin \theta } \right) = r\left\{ {\left[ {\hat i\cos \left( {\theta _0 + 2\pi } \right) + \hat j\sin \left( {\theta _0 + 2\pi } \right)} \right] - \left( {\hat i\cos \theta _0 + \hat j\sin \theta _0 } \right)} \right\} = 0
Thus, you will have
ZERO displacement.
And when you think about it, much of what I wrote is completely unnecessary! Though it does, "sort of" provide an answer from a mathematical perspective...err, though it might have been unnecessary.
sorry...
