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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 ... .
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