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.
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.
If the center is internal, the two geometric figures are scaled mirror images of one another; in technical language, they have opposite chirality.
Conversely, if the center is external, the two figures are directly similar to one another; their angles have the same sense.
Hence, a pair of circles has both types of homothetic centers, internal and external, unless the centers are equal or the radii are equal; these exceptional cases are treated after general position.
For a given pair of circles, the internal and external homothetic centers may be found in various ways.
More generally, taking both radii with the same sign (both positive or both negative) yields the inner center, while taking the radii with opposite signs (one positive and the other negative) yields the outer center.
Note that the equation for the inner center is valid for any values (unless both radii zero or one is the negative of the other), but the equation for the external center requires that the radii be different, otherwise it involves division by zero.
In synthetic geometry, two parallel diameters are drawn, one for each circle; these make the same angle α with the line of centers.
The lines A1A2, B1B2 drawn through corresponding endpoints of those radii, which are homologous points, intersect each other and the line of centers at the external homothetic center.
Conversely, the lines A1B2, B1A2 drawn through one endpoint and the opposite endpoint of its counterpart intersects each other and the line of centers at the internal homothetic center.
If the circles fall on opposite sides of the line, it passes through the internal homothetic center, as in A2B1 in the figure above.
Conversely, if the circles fall on the same side of the line, it passes through the external homothetic center (not pictured).
An external center can be defined in the projective plane to be the point at infinity corresponding to the slope of this line.
If one radius is zero but the other is non-zero (a point and a circle), both the external and internal center coincide with the point (center of the radius zero circle).
In general, a line passing through a homothetic center intersects each of its circles in two places.
Segment RQ' is seen in the same angle from P and S', which means R, P, S', Q' lie on a circle.
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.
Tangents drawn from the radical center to the three circles would all have equal length.
Consider the two rays emanating from the external homothetic center E in Figure 4.
The proof for the other pair of antihomologous points (P, Q'), as well as in the case of the internal homothetic center is analogous.
To show this, consider two rays from the homothetic center, intersecting the given circles (Figure 8).
As we've already shown these points lie on a circle C and thus the two rays are radical axes for C/T1, C/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
Thus the powers of E with respect to the two tangent circles are equal which means that E belongs to the radical axis.
Offset each center point perpendicularly to the plane by a distance equal to the corresponding radius.
The lines pierce the plane of circles in the points HAB, HBC, HAC.
This property is exploited in Joseph Diaz Gergonne's general solution to Apollonius' problem.