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 $\overline{OA} \cdot \overline{OA'} = k^2$ , 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

1. 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.

2. 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 $D$ .

3. 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: $r = \frac{k^2 \cdot r'}{\overline{OC_2}^2 - r'^2}.$

Problems that use Circular Inversion

Problem 1 (AMC12)

Source: 2017 AMC12A Problem 16

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 $P$ ?

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

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

Solution using Circular Inversion

Let $\Omega$ be a circle with radius of $6$ and centered at the left corner of the semi-circle (O) with radius $3$ . Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram:

$[asy] size(5cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(6, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label("$\Omega$", (6, 0) * dir(30), NE); label("$C_1$", (3, 0), N); label("$C_2$", (2, 0), N); label("$C_3$", (5, 0), N); label("$C_4$", P, N); label("$O$", origin, W); [/asy]$

$C_1$ has radius $3$ , $C_2$ has radius $2$ , and $C_3$ has radius $1$ . We want to find the radius of $C_4$ .

We now invert the four circles. $C_1$ inverts to a line. Given that one point is on $\Omega$ , and all points on $\Omega$ invert to themselves, we know that the resulting line must intersect that intersection point. $C_2$ also inverts to a line. $C_2$ has radius $4$ , and since $\Omega$ has radius of $6$ , the resulting line must be $\frac{36}{4} = 9$ units away from $O$ . $C_3$ inverts to a circle. By observing the diagram, we note that $C_3'$ 's center must be on $\overline{OC_3}$ and be between the two inverted lines, because $C_3$ is tangent to $C_1$ and $C_2$ (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius $\frac{3}{2}$ that is $\frac{15}{2}$ units from $O$ .

$[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label("$\Omega$", (6, 0) * dir(30), NW); label("$C_1$", (3, 0), N); label("$C_2$", (2, 0), N); label("$C_3$", (5, 0), N); label("$C_4$", P, N); label("$O$", origin, W); draw((6, 3)--(6, -3)); draw((9, 3)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label("$C_1'$", (6, 3), N); label("$C_2'$", (9, 3), N); label("$C_3'$", (15/2, 0), N); dot((15/2, 0)); [/asy]$

Now, we invert $C_4$ . Note that $C_4$ is tangent to the three other original circles. So, in the inversion, $C_4'$ must be tangent to the two lines and $C_3'$ . It is then quickly seen that $C_3'$ and $C_4'$ have the same radius: $\frac{3}{2}$ .

Now, we can determine the radius of $C_4$ using the formula $r = \frac{k^2 \cdot r'}{\overline{OC_2}^2 - r'^2}$ . $k^2 = 36$ , and $r' = \frac{3}{2}$ . $\overline{OC}$ is just the distance from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$ . Since $\overline{OC_3'} = \frac{15}{2}$ , we use Pythagoras to learn that $\overline{OC_4'}^2 = \left(\frac{15}{2}\right)^2 + 9$ . We do not take the square root because our relationship formula takes $\overline{OC_4'}^2$ .

Therefore, we have: $r = \frac{36 \cdot \frac{3}{2}}{\left(\frac{15}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 9} = \frac{54}{9 \cdot 6 + 9} = \frac{6}{7} = \boxed{B}.$

Here is the diagram with $C_4'$ .

$[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -45, 45); draw(inversion, dashed); label("$\Omega$", (6, 0) * dir(45), NW); label("$C_1$", (3, 0), N); label("$C_2$", (2, 0), N); label("$C_3$", (5, 0), N); label("$C_4$", P, N); label("$O$", origin, W); draw((6, 4.5)--(6, -3)); draw((9, 4.5)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label("$C_1'$", (6, -3), SW); label("$C_2'$", (9, -3), SE); label("$C_3'$", (15/2, 0), NE); dot((15/2, 0)); draw(Circle((15/2, 3), 3/2)); dot((15/2, 3)); label("$C_4'$", (15/2, 3), N); draw((15/2, 0)--(15/2, 3)); draw(rightanglemark(origin, (15/2, 0), (15/2, 3))); [/asy]$

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Circular Inversion

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