Homothetic center

Figure 1: The point O is an external homothetic center for the two triangles. The size of each figure is proportional to its distance from the homothetic center.

In geometry, a homothetic center (also called a center of similarity or a center of similitude) is a point from which at least two geometrically similar figures can be seen as a dilation/contraction of one another. If the center is external, the two figures are directly similar to one another; their angles have the same rotational sense. If the center is internal, the two figures are scaled mirror images of one another; their angles have the opposite sense.

Figure 2: Two geometric figures related by an external homothetic center S. The angles at corresponding points are the same and have the same sense; for example, the angles ABC and A'B'C' are both clockwise and equal in magnitude.

General polygons

If two geometric figures possess a homothetic center, they are similar to one another; in other words, they must have the same angles at corresponding points and differ only in their relative scaling. The homothetic center and the two figures need not lie in the same plane; they can be related by a projection from the homothetic center.

Homothetic centers may be external or internal. If the center is internal, the two geometric figures are scaled mirror images of one another; in technical language, they have opposite chirality. A clockwise angle in one figure would correspond to a counterclockwise angle in the other. Conversely, if the center is external, the two figures are directly similar to one another; their angles have the same sense.

Circles

Figure 3: Two circles have both types of homothetic centers, internal (I) and external (E). The radii of the circles (r₁ and r₂) are proportional to the distance (d) from each homothetic center. The points A₁ and A₂ are homologous, as are the points B₁ and B₂.

Circles are geometrically similar to one another and mirror symmetric. Hence, any pair of circles has both types of homothetic centers, internal and external; their two homothetic centers lie on the line joining the centers of the two given circles, which is called the line of centers (Figure 3).

For a given pair of circles, the internal and external homothetic centers may be found as follows. Two radii are drawn in the two circles such that they make the same angle with the line joining their centers (the angle α in Figure 3). A line drawn through corresponding endpoints of those radii (e.g., the points A₁ and A₂ in Figure 3) intersects the line of centers at the external homothetic center. Conversely, a line drawn through one endpoint and the diametric opposite endpoint of its counterpart (e.g., the points A₁ and B₂ in Figure 3) intersects the line of centers at the internal homothetic center.

As a limiting case of this construction, a line tangent to both circles, where the circles fall on opposite sides, passes through the internal homothetic center, as illustrated by the line joining the points B₁ and A₂ in Figure 3. Conversely, a line tangent to both given circle, where the circles fall on the same side, passes through the external homothetic center. See also Tangent lines to circles.

Homologous and antihomologous points

Figure 4: Lines through corresponding antihomologous points intersect on the radical axis of the two given circles (green and blue). The points Q and P′ are antihomologous, as are S and R′. These four points lie on a circle that intersects the two given circles; the lines through the intersection points of the new circle with the two given circles must intersect at the radical center G of the three circles, which lies on the radical axis of the two given circles.

In general, a ray emanating from a homothetic center will intersect each of its circles in two places. Of these four points, two are said to be homologous if radii drawn to them make the same angle with the line connecting the centers, e.g., the points A₁ and A₂ in Figure 3. Points which are collinear with respect to the homothetic center but are not homologous are said to be antihomologous, e.g., points Q and P′ in Figure 4.

Pairs of antihomologous points lie on a circle

When two rays from the same homothetic center intersect the circles, each set of antihomologous points lie on a circle.

Let's consider triangles EQS and EQ′S′ (Figure 4).
They are similar because both share angle ∠QES=∠Q′ES′ and $\frac{EQ}{EQ^\prime}=\frac{ES}{ES^\prime}$ since E is the homothetic center. From that similarity follows that ∠ESQ=∠ES′Q′=α. Because of the inscribed angle theorem ∠EP′R′=∠ES′Q′. ∠QSR′=180°-α since it is supplementary to ∠ESQ. In the quadrilateral QSR′P′ ∠QSR′+∠QP′R′=180°-α+α=180° which means it can be inscribed in a circle. From the secant theorem follows that EQ·EP′=ES·ER′.

In the same way it can be shown that PRS′Q′ can be inscribed in circle and EP·EQ′=ER·ES′.

The proof is similar for the internal homothetic center I.
PIR~P′IR′ then ∠RPI=∠IP′R′=α. ∠RS′Q′=∠PP′R′=α (inscribed angle theorem). Segment RQ′ is seen in the same angle from P and S′ which means R, P, S′ and Q′ lie on a circle. Then from intersecting chords theorem IP·IQ′=IR·IS′. Similarly QSP′R′ can be inscribed in a circle and IQ·IP′=IS·IR′.

Relation with the radical axis

Two circles have a radical axis, which is the line of points from which tangents to both circles have equal length. More generally, every point on the radical axis has the property that its powers relative to the circles are equal. The radical axis is always perpendicular to the line of centers, and if two circles intersect, their radical axis is the line joining their points of intersection. For three circles, three radical axes can be defined, one for each pair of circles (C₁/C₂, C₁/C₃, and C₂/C₃); remarkably, these three radical axes intersect at a single point, the radical center. Tangents drawn from the radical center to the three circles would all have equal length.

Any two pairs of antihomologous points can be used to find a point on the radical axis. Consider the two rays emanating from the external homothetic center E in Figure 4. These rays intersect the two given circles (green and blue in Figure 4) in two pairs of antihomologous points, Q and P′ for the first ray, and S and R′ for the second ray. These four points lie on a single circle, that intersects both given circles. By definition, the line QS is the radical axis of the new circle with the green given circle, whereas the line P′R′ is the radical axis of the new circle with the blue given circle. These two lines intersect at the point G, which is the radical center of the new circle and the two given circles. Therefore, the point G also lies on the radical axis of the two given circles.

Tangent circles and antihomologous points

For each pair of antihomologous points of two circles exists a third circle which is tangent to the given ones and touches them at the antihomologous points.
The opposite is also true — every circle which is tangent to two other circles touches them at a pair of antihomologous points.

Figure 5: Every circle which is tangent to two given circles touches them at a pair of antihomologous points

Let our two circles have centers O₁ and O₂ (Figure 5). E is their external homothetic center. We construct an arbitrary ray from E which intersects the two circles in P, Q, P′ and Q′. Extend O₁Q and O₂P′ until they intersect in T₁. It is easily proven that triangles O₁PQ and O₂P′Q′ are similar because of the homothety. They are also isosceles because O₁P=O₁Q (radius), therefore ∠O₁PQ=∠O₁QP=∠O₂P′Q′=∠O₂Q′P′=∠T₁QP′=∠T₁P′Q. Thus T₁P′Q is also isosceles and a circle can be constructed with center T₁ and radius T₁P′=T₁Q. This circle is tangent to the two given circles in points Q and P′.

The proof for the other pair of antihomologous points (P and Q′), as well as in the case of the internal homothetic center is analogous.

Figure 6: Family of tangent circles for the external homothetic center

Figure 7: Family of tangent circles for the internal homothetic center

If we construct the tangent circles for every possible pair of antihomologous points we get two families of circles - one for each homothetic center. The family of circles of the external homothetic center is such that every tangent circle either contains both given circles or none (Figure 6). On the other hand the circles from the other family always contain only one of the given circles (Figure 7).

Figure 8: The radical axis of tangent circles passes through the radical center

All circles from a tangent family have a common radical center and it coincides with the homothetic center.
To show this let's consider two rays from the homothetic center, intersecting the given circles (Figure 8). Two tangent circles T₁ and T₂ exist which touch the given circles at the antihomologous points. As we've already shown these points lie on a circle C and thus the two rays are radical axes for C/T₁ and C/T₂. Then the intersecting point of the two radical axes must also belong to the radical axis of T₁/T₂. This point of intersection is the homothetic center E.

If the two tangent circle touch collinear pairs of antihomologous point — as in Figure 5 — then because of the homothety$\frac{EP}{EP^\prime}=\frac{EQ}{EQ^\prime}; {EP}\cdot{EQ^\prime}={EQ}\cdot{EP^\prime}$. Thus the powers of E with respect to the two tangent circles are equal which means that E belongs to the radical axis.

Homothetic centers of three circles

See also: Monge's theorem

Any pair of circles has two centers of similarity, therefore, three circles would have six centers of similarity, two for each distinct pair of given circles. Remarkably, these six points lie on four lines, three points on each line. Here is one way to show this.

Figure 9: In a three circle configuration, three homothetic centers (one for each pair of circles) lie on a single line

Consider the plane of the three circles (Figure 9). Offset each center point perpendicularly to the plane by a distance equal to the corresponding radius. The centers can be offset to either side of the plane. The three offset points define a single plane. In that plane we build three lines through each pair of points. The lines pierce the plane of circles in the points H_AB, H_BC and H_AC. Since the locus of points which are common to two distinct and non-parallel planes is a line then necessarily these three points lie on such line. From the similarity of triangles H_ABAA′ and H_ABBB′ we see that $\frac{H_{AB}B}{H_{AB}A}=\frac{r_B}{r_A}$ (r_A,B being the radii of the circles) and thus H_AB is in fact the homothetic center of the corresponding two circles. We can do the same for H_BC and H_AC.

Figure 10: All six homothetic centers (dots) of three circles lie on four lines (thick lines)

Repeating the above procedure for different combinations of homothetic centers (in our method this is determined by the side to which we offset the centers of the circles) would yield a total of four lines — three homothetic centers on each line (Figure 10).

Here is yet another way to prove this.

Figure 11: The blue line is the radical axis of the two tangent circles C₁ and C₂ (pink). Each pair of given circles has a homothetic center which belongs to the radical axis of the two tangent circles. Since the radical axis is a line this means that the three homothetic centers are collinear

Let C₁ and C₂ be a conjugate pair of circles tangent to all three given circles (Figure 11). By conjugate we imply that both tangent circles belong to the same family with respect to any one of the given pairs of circles. As we've already seen, the radical axis of any two tangent circles from the same family passes through the homothetic center of the two given circles. Since the tangent circles are common for all three pairs of given circles then their homothetic centers all belong to the radical axis of C₁ and C₂ e.g., they lie on a single line.

This property is exploited in Joseph Diaz Gergonne's general solution to Apollonius' problem. Given the three circles, the homothetic centers can be found and thus the radical axis of a pair of solution circles. Of course, there are infinitely many circles with the same radical axis, so additional work is done to find out exactly which two circles are the solution.

See also

• Similarity (geometry)
• Homothetic transformation
• Radical axis, radical center
• Apollonius' problem

References

• Johnson RA (1960). Advanced Euclidean Geometry: An Elementary treatise on the geometry of the Triangle and the Circle. New York: Dover Publications. 
• Kunkel, Paul (2007), "The tangency problem of Apollonius: three looks", BSHM Bulletin: Journal of the British Society for the History of Mathematics 22 (1): 34–46, doi:10.1080/17498430601148911, http://whistleralley.com/tangents/bshmkunkel.pdf