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Circular Inversion

Revision as of 19:53, 30 March 2017 by Pandabear10 (talk | contribs)

Note: Page is under construction

Circular Inversion, sometimes called Geometric Inversion, is a transformation where point $A$ in the Cartesian plane is transformed based on a circle $\Omega$ with radius $k$ and center $O$ such that $\overline{OA} \times \overline{OA'} = k^2$, where $A'$ is the transformed point on the ray extending from $O$ through $A$.

[asy] size(5cm); draw(Circle((0, 0), 9), dashed); pair O = (0, 0); dot(O); label("$O$", O, NW); pair A = (6, 0); dot(A); label("$A$", A, NW); pair B = (27/2, 0); dot(B); label("$A'$", B, N); draw(O--B); draw(O--(0, 9)); label("$k$", O--(0, 9), W); label("$\Omega$", (-9, 9), SE); [/asy]

Note that $A'$, when inverted, transforms back to $A$. All points outside of $\Omega$ are transformed inside $\Omega$, and vice versa. Points on $\Omega$ transform to themselves, meaning $A = A'$. Finally, the transformation of $O$ is debated on its existence. Some call the transformation the ideal point, which is infinitely far away and in every direction. Others claim that this point does not have an inverse.

Geometric Inversion technically refers to many different types of inversions, however, if Geometric Inversion is used without clarification, Circular Inversion is usually assumed.

Circular Inversion can be a very useful tool in solving problems involving many tangent circles and/or lines.

Basics of Circular Inversion

Inversion of a Circle intersecting O

[asy] unitsize(10); size(5cm); draw(Circle((0, 0), 9), dashed); pair O = (0, 0); dot(O); label("$O$", O, NW); pair P = (6, 0); dot(P); label("$P$", P, SE); pair A = (27/2, 0); dot(A); draw(O--A); label("$P'$", A, NE); pair B = (27/2, 8); dot(B); label("$Q'$", B, NE); draw(O--B); path circle = Circle((3, 0), 3); path line = Line(O, B); pair[] x = intersectionpoints(circle, line); pair Q = x[0]; dot(Q); label("$Q$", Q, N); draw(circle); draw(O--(0, 9)); label("$k$", O--(0, 9), W); label("$\Omega$", (-8, 8), SE); draw(A--B); draw(Q--P); pair C = (3, 0); dot(C); label("$C$", C, S); [/asy]

The first thing that we must learn about inversion is what happens when a circle which intersects the center of the inversion, $O$, is inverted. Let us have circle $C$, with diameter $\overline{OP}$. $Q$ is chosen arbitrarily on circle $C$. Points $P'$ and $Q'$ represent the inversions of $P$ and $Q$, respectively. $k$ is the radius of $\Omega$. We seek to show that circle $C$ inverts to a line perpendicular to $\overline{OP}$ through $P'$.

By the definition of inversion, we have $\overline{OP} \times \overline{OP'} = k^2$ and $\overline{OQ} \times \overline{OQ'}$.

We can combine the two equations to get $\overline{OP} \times \overline{OP'} = \overline{OQ} \times \overline{OQ'}$. Rewriting this gives: \[\frac{\overline{OP}}{\overline{OQ}} = \frac{\overline{OQ'}}{\overline{OP'}}.\]

Also, since $\overline{OP}$ is a diameter of circle $C$, $\angle OQP$ must be right.

Now, we consider $\triangle OQP$ and $\triangle OP'Q'$. They share an angle - $\angle QOP$, and we know that $\frac{\overline{OP}}{\overline{OQ}} = \frac{\overline{OQ'}}{\overline{OP'}}.$ Therefore, we have SAS similarity. Therefore, $\angle{OP'Q'}$ must be right. From there, it follows that all points on circle $C$ will be inverted onto the line perpendicular to $\overline{OP}$ at $P$.

Therefore, the inversion of circle $C$ becomes a line.

Note that, if circle $C$ extends beyond $\Omega$, the argument still holds. All one needs to do is shuffle things around.

Inversion of a Circle not intersecting O

[asy] unitsize(7); int radiusInverse = 12; int k = radiusInverse * radiusInverse; int radiusCircle = 4; int circleCenterLocation = 7; path inverse = Circle((0, 0), radiusInverse); draw(inverse, dashed); label("$\Omega$", inverse, NW); pair O = (0, 0); dot(O); label("$O$", O, NW); pair C = (circleCenterLocation, 0); dot(C); label("$C$", C, S); path circle1 = Circle(C, radiusCircle); draw(circle1); int p = circleCenterLocation - radiusCircle; int q = circleCenterLocation + radiusCircle; pair P = (p, 0); pair Q = (q, 0); pair A = (k / p, 0); pair B = (k / q, 0); pair D = (((k / p) + (k / q)) / 2, 0); dot(P); dot(Q); dot(A); dot(B); label("$P$", P, SE); label("$Q$", Q, SW); label("$Q'$", B, SE); label("$P'$", A, SE); draw(O--A); path circle2 = Circle(D, ((k / p) - (k / q)) / 2); //draw(circle2); path invisibleLine = Line(O, (48, 18)); pair[] x = intersectionpoints(circle1, invisibleLine); pair[] y = intersectionpoints(circle2, invisibleLine); pair R = x[0]; pair S = y[1]; dot(R); dot(S); label("$R$", R, N); label("$R'$", S, NW); draw(O--S); draw(P--R--Q); draw(B--S--A); markscalefactor=0.1; draw(rightanglemark(P, R, Q)); [/asy]

Now, we study the inversion of a circle not intersecting the center of inversion. Let us have circle $C$ not intersecting $O$, the center of $\Omega$, the circle which we invert around. The points where $\overline{OC}$ intersect circle $C$ are points $P$ and $Q$, respectively. Point $R$ is arbitrary and on circle $C$. We invert points $P$, $Q$, and $R$, producing $P'$, $Q'$, and $R'$, respectively. We draw $\triangle PRQ$ and $\triangle OR'P'$. Because $\overline{PQ}$ is a diameter, $\angle PRQ$ must be right. We wish to show that circle $C$ inverts to another circle.

The definition of inversion tells us that $\overline{OP} \times \overline{OP'} = \overline{OQ} \times \overline{OQ'} = \overline{OR} \times \overline{OR'} = k^2$. From here, we obtain that $\frac{\overline{OP}}{\overline{OR}} = \frac{\overline{OR'}}{\overline{OP'}}$ and $\frac{\overline{OQ}}{\overline{OR}} = \frac{\overline{OR'}}{\overline{OQ'}}$ By SAS symmetry (exploiting $\angle ROQ$), the ratios tell us that: \[\triangle ORP \sim \triangle OP'R', \triangle ORQ \sim \triangle OQ'R'.\]

Therefore, we have $\angle OPR = \angle OR'P'$ and $\angle OQR = \angle OR'Q'$. Note that $\angle OPR = \angle PRQ + \angle PQR = 90^\circ + \angle PQR$, which must equal $\angle OR'P'$. Therefore, $\angle Q'R'P' = \angle OR'P' - \angle OR'Q' = 90^\circ + \angle PQR - \angle OQR$. But $\angle OQR = \angle PQR$. Therefore, $\angle Q'R'P' = 90^\circ$. As this holds for any $R$, all points on circle $C$ will invert to a point on a circle with diameter $D$.

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