Given a point Z and an inversion circle

with center at point Q and radius

, consider the following construction to obtain the inverse Z' of Z in

. Draw the circle centered at Z passing through Q; let it intersect

at points A and B. Draw two circles

and

centered at A and B both passing through Q. Their second point of intersection is the inverse of Z in

. We justify this by the following reasoning: if we invert

and

in

we obtain (red dotted) lines passing through Z; these lines intersect at

and Z, hence their inverses must intersect at the inverses of

and Z, namely Q and Z'. This Demonstration lets you drag the point Z (red); the construction works for all points Z at a distance from Q greater than

.