The diagram shows the trace of Q and its image, as Q moves from B towards D.

We know the angle at B is a right angle (because BD was chosen to be perpendicular to OB).
It turns out that the angle at Q' is also a right angle. One way to prove this is to show that triangles OBQ and OQ'B' are similar. We can do this by using the property of the inversion transformation, that OBxOB' = OQxOQ'. Can you see how?

In turn, we can use our knowledge of circle theorems to prove that the image of Q lies on a circle with diameter OB'.

On the next page we look at the image of a point that moves on a circle that passes through O.