# Constructing the inverse of a point with respect to a circle

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**Definition.** Consider a circle with centre \(O\) and radius \(r\). Given any point \(P\) in the plane, the *inverse* of \(P\) with respect to the circle is the unique point \(P'\) on the ray \(OP\) such that \(|OP| |OP'| = r^2\).

## Constructing the inverse using tangent lines

The first picture below illustrates how to construct the inverse of a point \(P\) using tangent lines. Draw a line through \(P\) which is orthogonal to the ray \(OP\). Let \(X\) and \(Y\) be the two points where this line intersects the circle. Now draw tangents to the circle at \(X\) and \(Y\) and let \(P'\) be the point where these tangents intersect. Then \(P'\) is the inverse of \(P\).

## Angle bisector construction of the inverse

The next picture explains a different construction of the inverse. The circle centred at \(O\) and the point \(P\) are given, as well as a point \(X\) on the circle. Let \(Z\) be the intersection of the circle and the ray \(OP\). Choose a point \(P'\) on the ray \(OP\) such that \(XZ\) is the angle bisector of \(\angle PXP'\). Then \(P'\) is the inverse of \(P\) with respect to the circle.

## Isosceles triangle construction of the inverse

A third construction of the inverse uses isosceles triangles. In the diagram below, the circle is centred at \(O\), the point \(P\) is given and we want to construct the inverse \(P'\) of \(P\). Let \(X\) be the intersection of the circle with the perpendicular bisector of \(OP\). Now let \(P'\) be the intersection of the ray \(OP\) with the perpendicular bisector of \(OX\). Then \(P'\) is the inverse of \(P\).

## All three constructions

In the final diagram, we see all three constructions at once. The lines in the tangent construction are coloured orange, the angle bisector construction is coloured red and the isosceles triangle construction is coloured green. You can move the points \(O\), \(P\) and \(R\) around to convince yourself that all three constructions do indeed give the same point \(P'\).