You will have completed a circle when ... come to think of it, this actually involves some math involving the wheel diameter, the axle length, and the rate at which the two servos are rotating.

After completing a circle, the circumference of the circle traced out by the outer wheel will be 2πl greater than that traced out by the inner one (where l is the axle length). So if the outer wheel is travelling at Vo and the inner one at Vi, you will get a complete circle at 2πl/(Vo-Vi) seconds. Vo is πd omega_o, where d is the wheel diameter and omega_o is the rotational speed of the outer wheel in rotations per second. So we get 2πl / [πd (omega_o - omega_i)]. The π cancels out a little suspiciously, leaving us with 2l/d * 1/(difference of the wheel speeds in rotations per second) seconds to complete a circle.

Something like that, anyway. I shall leave the calculation of the diameter of the circle in which the outer wheel travels as an exercise for the reader. Hint: you want this to be smaller than the floor.

In any case, to do a figure 8: go left, wait till you have finished going left, then go right and wait untill you have finished going right. People omit that last step, and wonder why their robot only goes left.