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# 10.5: Perpendicular circles

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Assume two circles $$\Gamma$$ and $$\Omega$$ intersect at two points $$X$$ and $$Y$$. Let $$\ell$$ and $$m$$ be the tangent lines at $$X$$ to $$\Gamma$$ and $$\Omega$$ respectively. Analogously, $$\ell'$$ and $$m'$$ be the tangent lines at $$Y$$ to $$\Gamma$$ and $$\Omega$$.

From Exercise 9.6.3, we get that $$\ell \perp m$$ if and only if $$\ell' \perp m'$$.

We say that the circle $$\Gamma$$ is perpendicular to the circle $$\Omega$$ (briefly $$\Gamma \perp \Omega$$) if they intersect and the lines tangent to the circles at one point (and therefore, both points) of intersection are perpendicular.

Similarly, we say that the circle $$\Gamma$$ is perpendicular to the line $$\ell$$ (briefly $$\Gamma \perp \ell$$) if $$\Gamma \cap \ell \ne \emptyset$$ and $$\ell$$ perpendicular to the tangent lines to $$\Gamma$$ at one point (and therefore, both points) of intersection. According to Lemma 5.6.2, it happens only if the line l passes thru the center of $$\Gamma$$.

Now we can talk about perpendicular circlines.

Theorem $$\PageIndex{1}$$

Assume $$\Gamma$$ and $$Omega$$ are distinct circles. Then $$\Omega \perp \Gamma$$ if and only if the circle $$\Gamma$$ coincides with its inversion in $$\Omega$$.

Proof

Suppose that $$\Gamma'$$ denotes the inverse of $$\Gamma$$. "Only if" part. Let $$O$$ be the center of $$\Omega$$ and $$Q$$ be the center of $$\Gamma$$. Let $$X$$ and $$Y$$ denote the points of intersections of $$\Gamma$$ and $$Omega$$. According to Lemma 5.6.2, $$\Omega \perp \Gamma$$ if and only if $$(OX)$$ and $$(OY)$$ are tangent to $$\Gamma$$.

Note that $$\Gamma'$$ is also tangent to $$(OX)$$ and $$(OY)$$ and $$X$$ and $$Y$$ respectively. It follows that $$X$$ and $$Y$$ are the foot points of the center of $$\Gamma'$$ on $$(OX)$$ and $$(OY)$$. Therefore, both $$\Gamma'$$ and $$\Gamma$$ have the center $$Q$$. Finally, $$\Gamma' = \Gamma$$, since both circles pass thru $$X$$.

"If" part. Assume $$\Gamma = \Gamma'$$.

Since $$\Gamma \ne \Omega$$, there is a point $$P$$ that lies on $$\Gamma$$, but not on $$\Omega$$. Let $$P'$$ be the inverse of $$P$$ in $$\Omega$$. Since $$\Gamma = \Gamma'$$, we have that $$P' \in \Gamma$$. In particular, the half-line $$[OP)$$ intersects $$\Gamma$$ at two points. By Exercise 5.6.1, $$O$$ lies outside of $$\Gamma$$.

As $$\Gamma$$ has points inside and outside of $$\Omega$$, the circles $$\Gamma$$ and $$\Omega$$ intersect. The latter follows from Exercise 3.5.1.

Let $$X$$ be point of their intersection. We need to show that $$(OX)$$ is tangent to $$\Gamma$$; that is, $$X$$ is the only intersection point of $$(OX)$$ and $$\Gamma$$.

Assume $$Z$$ is another point of intersection. Since $$O$$ is outside of $$\Gamma$$, the point $$Z$$ lies on the half-line $$[OX)$$.

Suppose that $$Z'$$ denotes the inverse of $$Z$$ in $$\Omega$$. Clearly, the three points $$Z$$, $$Z'$$, $$X$$ lie on $$\Gamma$$ and $$(OX)$$. The latter contradicts Lemma 5.6.1.

It is convenient to define the inversion in the line $$\ell$$ as the reflection across $$\ell$$. This way we can talk about inversion in an arbitrary circline.

Corollary $$\PageIndex{1}$$

Let $$\Omega$$ and $$\Gamma$$ be distinct circlines in the inversive plane. Then the inversion in $$\Omega$$ sends $$\Gamma$$ to itself if and only if $$\Omega \perp \Gamma$$.

Proof

By Thorem $$\PageIndex{1}$$, it is sufficient to consider the case when $$\Omega$$ or $$\Gamma$$ is a line.

Assume $$\Omega$$ is a line, so the inversion in $$\Omega$$ is a reflection. In this case the statement follows from Corollary 5.4.1.

If $$\Gamma$$ is a line, then the statement follows from Theorem 10.3.2.

Corollary $$\PageIndex{2}$$

Let $$P$$ and $$P'$$ be two distinct points such that $$P'$$ is the inverse of $$P$$ in the circle $$\Omega$$. Assume that the circline $$\Gamma$$ passes thru $$P$$ and $$P'$$. Then $$\Gamma \perp \Omega$$.

Proof

Without loss of generality, we may assume that $$P$$ is inside and $$P'$$ is outside $$\Omega$$. By Theorem 3.5.1, $$\Gamma$$ intersects $$\Omega$$. Suppose that A denotes a point of intersection.

Suppose that $$\Gamma'$$ denotes the inverse of $$\Gamma$$. Since $$A$$ is a self-inverse, the points $$A, P$$, and $$P'$$ lie on $$\Gamma'$$. By Exercise 8.1.1, $$\Gamma'$$ = $$\Gamma$$ and by Theorem $$\PageIndex{1}$$, $$\Gamma \perp \Omega$$.

Corollary $$\PageIndex{3}$$

Let $$P$$ and $$Q$$ be two distinct points inside the circle $$\Omega$$. Then there is a unique circline $$\Gamma$$ perpendicular to $$\Omega$$ that passes thru $$P$$ and $$Q$$.

Proof

Let $$P'$$ be the inverse of the point $$P$$ in the circle $$\Omega$$. According to Corollary $$\PageIndex{2}$$, the circline is passing thru $$P$$ and $$Q$$ is perpendicular to $$\Omega$$ if and only if it passes thru $$P'$$.

Note that $$P'$$ lies outside of $$\Omega$$. Therefore, the points $$P$$, $$P'$$, and $$Q$$ are distinct.

According to Exercise Exercise 8.1.1, there is a unique circline passing thru $$P, Q$$, and $$P'$$. Hence the result.

Exercise $$\PageIndex{1}$$

Let $$P, Q, P'$$, and $$Q'$$ be points in the Euclidean plane. Assume $$P'$$ and $$Q'$$ are inverses of $$P$$ and $$Q$$ respectively. Show that the quadrangle $$PQP'Q'$$ is inscribed.

Hint

Apply Theorem 10.2.1, Theorem 7.4.5 and Theorem 9.2.1.

Exercise $$\PageIndex{2}$$

Let $$\Omega_1$$ and $$\Omega_2$$ be two perpendicular circles with centers at $$O_1$$ and $$O_2$$ respectively. Show that the inverse of $$O_1$$ in $$\Omega_2$$ coincides with the inverse of $$O_2$$ in $$\Omega_1$$.

Hint

Suppose that $$T$$ denotes a point of intersection of $$\Omega_1$$ and $$\Omega_2$$. Let $$P$$ be the foot point of $$T$$ on $$(O_1O_2)$$. Show that $$\triangle O_1PT \sim \triangle O_1TO_2 \sim \triangle TPO_2$$. Consider that $$P$$ coincides with the inverses of $$O_1$$ in $$\Omega_2$$ and of $$O_2$$ in $$\Omega_1$$.

Exercise $$\PageIndex{3}$$

Three distinct circles — $$\Omega_1$$, $$\Omega_2$$ and $$\Omega_3$$ , intersect at two points — $$A$$ and $$B$$. Assume that a circle $$\Gamma$$ is perpendicular to $$\Omega_1$$ and $$\Omega_2$$. Show that $$\Gamma \perp \Omega_3$$.

Hint

Since $$\Gamma \perp \Omega_1$$ and $$\Gamma \perp \Omega_2$$, Corollary $$PageIndex{1}$$ implies that the circles $$\Omega_1$$ and $$\Omega_2$$ are inverted in $$\Gamma$$ to themselves. Conclude that the points $$A$$ and $$B$$ are inverse of each other. Since $$\Omega_3 \ni A, B$$, Corollary $$\PageIndex{2}$$ implies that $$\Omega_3 \perp \Gamma$$.

Let us consider two new construction tools: the circumtool that constructs a circline thru three given points, and the inversion-tool — a tool that constructs an inverse of a given point in a given circline.

Exercise $$\PageIndex{4}$$

Given two circles $$\Omega_1$$, $$\Omega_2$$ and a point $$P$$ that does not lie on the circles, use only circum-tool and inversion-tool to construct a circline $$\Gamma$$ that passes thru $$P$$, and perpendicular to both $$\Omega_1$$ and $$\Omega_2$$.

Hint

Let $$P_1$$ and $$P_2$$ be the inverse of $$P$$ in $$\Omega_1$$ and $$\Omega_2$$. Apply Corollary $$\PageIndex{2}$$ and Theorem $$\PageIndex{1}$$ to show that a circline $$\Gamma$$ that pass thru

$$P, P_1$$, and $$P_3$$ is a solution.

Advanced Exercise $$\PageIndex{5}$$

Given three disjoint circles $$\Omega_1$$, $$\Omega_2$$ and $$\Omega_3$$, use only circum-tool and inversion-tool to construct a circline $$\Gamma$$ that perpendicular to each circle $$\Omega_1$$, $$\Omega_2$$ and $$\Omega_3$$.

Think what to do if two of the circles intersect.

Hint

All circles that perpendicular to $$\Omega_1$$ and $$\Omega_2$$ pass thru a fixed point $$P$$. Try to construct $$P$$.

If two of the circles intersect, try to apply Corollary 10.6.1.