
# 10.1: Inversion


Let $$\Omega$$ be the circle with center $$O$$ and radius $$r$$. The inversion of a point $$P$$ in $$\Omega$$ is the point $$P' \in [OP)$$ such that

$$OP \cdot OP' = r^2.$$

In this case the circle $$\Omega$$ will be called the circle of inversion and its center $$O$$ is called the center of inversion.

The inverse of $$O$$ is undefined.

Note that if $$P$$ is inside $$\Omega$$, then $$P'$$ is outside and the other way around. Further, $$P = P'$$ if and only if $$P \in \Omega$$.

Note that the inversion maps $$P'$$ back to $$P$$.

Exercise $$\PageIndex{1}$$

Let $$\Omega$$ be a circle centered at $$O$$. Suppose that a line $$(PT)$$ is tangent to $$\Omega$$ at $$T$$. Let $$P'$$ be the foot point of $$T$$ on $$(OP)$$.

Show that $$P'$$ is the inverse of $$P$$ in $$\Omega$$.

Hint

By Lemma 5.6.2, $$\angle OTP'$$ is right. Therefore, $$\triangle OPT \sim \triangle OTP'$$ and in particular $$OP \cdot OP' = OT^2$$ and hence the result.

Lemma $$\PageIndex{1}$$

Let $$\Gamma$$ be a circle with the center $$O$$. Assume $$A'$$ and $$B'$$ are the inverses of $$A$$ and $$B$$ in $$\Gamma$$. Then

$$\triangle OAB \sim \triangle OB'A'.$$

Moreover

$\begin{array} {rcl} {\measuredangle AOB} & \equiv & {-\measuredangle B'OA',} \\ {\measuredangle OBA} & \equiv & {-\measuredangle OA'B',} \\ {\measuredangle BAO} & \equiv & {-\measuredangle A'B'O.} \end{array}$

Proof

Let $$r$$ be the radius of the circle of the inversion.

By the definition of inversion,

$$OA \cdot OA' = OB \cdot OB' = r^2.$$

Therefore,

$$\dfrac{OA}{OB'} = \dfrac{OB}{OA'}.$$

Clearly,

$$\measuredangle AOB = \measuredangle A'OB' \equiv -\measuredangle B'OA'.$$

From SAS, we get that

$$\triangle OAB \sim \triangle OB'A'.$$

Applying Theorem 3.3.1 and 10.1.2, we get 10.1.1.

Exercise $$\PageIndex{2}$$

Let $$P'$$ be the inverse of $$P$$ in the circle $$\Gamma$$. Assume that $$P \ne P'$$. Show that the value $$\dfrac{PX}{P'X}$$ is the same for all $$X \in \Gamma$$.

The converse to the exercise above also holds. Namely, given a positive real number $$k \ne 1$$ and two distinct points $$P$$ and $$P'$$ the locus of points $$X$$ such that $$\dfrac{PX}{P'X} = k$$ forms a circle which is called the Apollonian circle. In this case $$P'$$ is inverse of $$P$$ in the Apollonian circle.

Hint

Suppose that $$O$$ denotes the center of $$\Gamma$$. Assume that $$X, Y \in \Gamma$$; in particular, $$OX = OY$$.

Note that the inversion sends $$X$$ and $$Y$$ to themselves. By Lemma $$\PageIndex{1}$$,

$$\triangle OPX \sim \triangle OXP'$$     and     $$\triangle OPY \sim \triangle OYP'$$.

Therefore, $$\dfrac{PX}{P'X} = \dfrac{OP}{OX} = \dfrac{OP}{OY} = \dfrac{PY}{P'Y}$$ and hence the result.

Exercise $$\PageIndex{3}$$

Let $$A',B'$$, and $$C'$$ be the images of $$A,B$$, and $$C$$ under the inversion in the cincircle of $$\triangle ABC$$. Show that the incenter of $$\triangle ABC$$ is the orthocenter of $$\triangle A'B'C'$$.

Hint

By Lemma $$\PageIndex{1}$$,

$$\measuredangle IA'B' \equiv -\measuredangle IBA$$,         $$\measuredangle IB'C' \equiv -\measuredangle ICB$$,         $$\measuredangle IC'A' \equiv -\measuredangle IAC$$,
$$\measuredangle IB'A' \equiv -\measuredangle IAB$$,         $$\measuredangle IC'B' \equiv -\measuredangle IBC$$,         $$\measuredangle IA'C' \equiv -\measuredangle ICA$$.

It remains to apply the theorem on the sum of angles of triangle (Theorem 7.4.1) to show that $$(A'I) \perp (B'C')$$, $$(B'I) \perp (C'A')$$ and $$(C'I) \perp (B'A')$$.

Exercise $$\PageIndex{4}$$

Make a ruler-and-compass construction of the inverse of a given point in a given circle.

Hint

Guess the construction from the diagram (the two nonintersecting lines on the diagram are parallel).