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 .