Latest revision as of 17:49, 15 February 2025

Circular Inversion, sometimes called Geometric Inversion or simply 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 $\overline{OA} \cdot \overline{OA'} = k^2$ , 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 $\overline{OP} \cdot \overline{OP'} = k^2$ and $\overline{OQ} \cdot \overline{OQ'} = k^2$ .

We can combine the two equations to get $\overline{OP} \cdot \overline{OP'} = \overline{OQ} \cdot \overline{OQ'}$ . 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'$", A, SE); draw(O--A); path circle2 = Circle(D, ((k / p) - (k / q)) / 2); path invisibleLine = Line(O, (48, 18)); pair[] x = intersectionpoints(circle1, invisibleLine); pair[] y = intersectionpoints(circle2, invisibleLine); pair R = x[0]; pair S = y[1]; dot(R); dot(S); label("$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 $\overline{OP} \cdot \overline{OP'} = \overline{OQ} \cdot \overline{OQ'} = \overline{OR} \cdot\overline{OR'} = k^2$ . 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 $\overline{Q'P'}$ .

General Formula for the Radius of a Circle in Terms of the Radius of its Inverse Circle

$[asy] unitsize(7); int radiusInverse = 12; int k = radiusInverse * radiusInverse; int radiusCircle = 4; int circleCenterLocation = 7; path inverse = Circle((0, 0), radiusInverse); draw(inverse, dashed); label("$\Omega$", inverse, NW); pair O = (0, 0); dot(O); label("$O$", O, NW); pair C = (circleCenterLocation, 0); dot(C); label("$C_1$", C, S); path circle1 = Circle(C, radiusCircle); draw(circle1); int p = circleCenterLocation - radiusCircle; int q = circleCenterLocation + radiusCircle; pair P = (p, 0); pair Q = (q, 0); pair A = (k / p, 0); pair B = (k / q, 0); pair D = (((k / p) + (k / q)) / 2, 0); dot(P); dot(Q); dot(A); dot(B); dot(D); label("$C_2$", D, S); label("$P$", P, SE); label("$Q$", Q, SW); label("$Q'$", B, SE); label("$P'$", A, SE); draw(O--A); path circle2 = Circle(D, ((k / p) - (k / q)) / 2); draw(circle2); pair Tangent = (28, 20); //path tangentLine = tangent(O, D, (((k / p) - (k / q)) / 2), int n=1); path invisibleLine = Line(O, (28.723, 20)); pair[] x = intersectionpoints(circle1, invisibleLine); pair[] y = intersectionpoints(circle2, invisibleLine); pair R = x[0]; pair S = y[1]; dot(R); dot(S); label("$R$", R, N); label("$R'$", S, NW); draw(O--S--D); markscalefactor=0.1; draw(rightanglemark(O, S, D)); label("$r$", P--C, N); label("$r'$", B--D, N); draw(O--(0, radiusInverse)); label("$k$", O--(0, radiusInverse), W); [/asy]$

This is how circular inversion is useful in the first place - we find the radius of an inverted circle to find the radius of the original circle. Let the original circle be $C_1$ and the inverted circle be $C_2$ , with radii of $r$ and $r'$ , respectively. The radius of the circle of inversion is $k$ . We draw the tangent line of circle $C_1$ intersecting O. We know that this is also a tangent line to circle $C_2$ from the result from part 2 - the tangent line, by definition, intersects circle $C_1$ at exactly one point, and for every intersection point, part 2 says that there will be another intersection point. Therefore, the tangent line to circle $C_1$ intersects circle $C_2$ at exactly one point, necessitating this line to be a tangent line. Call the intersections $R$ and $R'$ , respectively. We have $\angle OR'C_2 = 90^\circ$ .

We have $\overline{OQ} - \overline{OP} = 2r$ and $\overline{OP'} - \overline{OQ'}$ = $2r'$ . We can write an equation for $\frac{r}{r'}$ by dividing: $\frac{r}{r'} = \frac{\overline{OQ} - \overline{OP}}{\overline{OP'} - \overline{OQ'}}.$

From the definition of inversion, we have $\overline{OQ}=\frac{k^2}{\overline{OQ'}}, \overline{OP}=\frac{k^2}{\overline{OP'}}$ . Subsituting yields: $\frac{r}{r'} = \frac{\frac{k^2}{\overline{OQ'}} - \frac{k^2}{\overline{OP'}}}{\overline{OP'} - \overline{OQ'}} = \frac{k^2\frac{\overline{OP'} - \overline{OQ'}}{\overline{OP'}\cdot\overline{OQ'}}}{\overline{OP'} - \overline{OQ'}} = \frac{k^2}{\overline{OP'}\cdot\overline{OQ'}}.$

From Power of a Point, we know that $\overline{OP'}\cdot\overline{OQ'} = \overline{OR'}^2$ , which equals $\overline{OC_2}^2 - r'^2$ . Subsistuting gives $\frac{r}{r'} = \frac{k^2}{\overline{OC_2}^2 - r'^2}$ , and solving for $r$ gives: $r = \frac{k^2 \cdot r'}{\overline{OC_2}^2 - r'^2}.$

Alternately, $\frac{r}{r'} = \frac{\overline{OR}}{\overline{OR'}} = \frac{k^2}{\overline{OR'}^2} = \frac{k^2}{\overline{OC_2}^2 - r'^2}.$

Problems

Problem 1

In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$ . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$ ?

$[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (-1,0)+(2+6/7)*dir(36.86989); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot(P); [/asy]$

$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{6}{7} \qquad\textbf{(C)}\ \frac{1}{2}\sqrt{3} \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} \qquad\textbf{(E)}\ \frac{11}{12}$

(Source)

This article is a stub. Help us out by expanding it.

Page

Toolbox

Search

Difference between revisions of "Circular Inversion"