Properties of the Circle of Similitude: What Is This about?
A Mathematical Droodle
|What if applet does not run?|
The circle of similitude of two circles of different radii that lie without each other exhibits several interesting properties, three of which are demonstrated by the applet below.
|What if applet does not run?|
With the reference to the applet, the circles c1, c2 are given by their centers O1, O2 and radii R1, R2. The centers of similarity of c1 and c2 are denoted P (internal) and Q (external). CM and CS are centers of segments O1O2 and PQ, respectively.
The circle of similitude (cS) of c1 and c2 is an Apollonian circle with respect two the centers O1 and O2 and ratio R1/R2 (or R2/R1). The circle is centered at CS and had PQ as a diameter. As such it belongs to a family of coaxal circles defined by the point circles O1, O2. Each member of the family, cS in particular, is orthogonal to the circle with O1O2 as a diameter. This is statement 1 hinted at by the applet.
The circle of similitude of any two circle with centers at O1, O2 is orthogonal to the circle with O1O2 as a diameter.
For a point T on the circle of similitude of two circles c1 and c2, the four points of contact of the tangents from T to the circles can be split into two pairs L1, L2 and M1, M2, with Li, Mi lying on ci,
|(2)||∠L1TP = ∠PTL2.|
For a point T on the circle of similitude, the tangents two the two circles form similar triangles. This makes angles L1TM1 and L2TM2 equal. The lines TO1 and TO2 serve naturally is the bisectors of these angles. The points of tangency L1, M1 for c1, and L2, M2 for c2, can be paired -- L1, L2 and M1, M2 -- so that the angles PTL1 and PTL2 are equal as are the angles PTM1 and PTM2. (The applet only shows only one pair of angles at a time.) This is because for an Apollonian circle CS, TP is a bisector of angle O1TO2. This proves statement 2.
Unlike Statements 1 and 2 that underscore equalities in angular relationships, Statements 3 and 4 relate to equalities of linear elements of the configuration.
The points in the pair L1 and L2 are equidistant from the radical axis of c1 and c2, and so are M1 and M2. (This is the same as saying that the midpoint of L1L2 (and also that of M1M2) lies on the radical axis.) If Hi,
|(3)||L1H1 = L2H2.|
Let the line L1L2 crosses c1 in K1 and c2 in K2. Then
|(4)||L1K1 = L2K2.|
The converse is also true: If a line cuts two circles in four points K1, L1, L2, K2 (in that order) so that (4) holds, then the tangents to the circles at L1 and L2 (as well as those at K1 and K2) meet on the circle of similitude.
Let t(X, c) denote the length of the tangential segment from a point X to a circle c. Then
|(4')||t(L1, c2) = t(L2, c1).|
Statements 4 and 4' are equivalent. Indeed,
Statements 3 and 4 are equivalent. The power of a point lying on a circle with respect to another circle is a linear function (see (5) below) of its distance to the radical axis of the two circles. Thus (3) implies (4') which, in turn, implies (4). The converse is as immediate.
Let F denote the intersection of the radical axis of the two circles c1, c2 with their line of centers. For a point S, let H be the projection of S on the latter. di,
the latter because the powers of F with respect to the two circles c1, c2 are equal, as are the powers of any other point on the radical axis of the circles. If d is the distance between the centers O1, O2, we can write
|Pow(S, c1) - Pow(S, c2) = 2·d·FH.|
For a point on c2,
|(5)||Pow(S, c1) = 2·d·FH.|
A similar formula for a point on c1 seems to have come with a sign minus, but this only because we have to consider FH as a signed segment. Of course all other segments in this discussion had be signed. This was not important until now.
Proof of Statement 4
(I thank Stuart Anderson for help with this proof.)
Let B be the foot of perpendicular from T to L1L2. (The following applies equally well to points M which, for this reason, won't be mentioned again.) Let Bi,
From the aforementioned similarities we obtain the following proportions:
L1B1 / R1 = TB / L1T and|
L2B2 / R2 = TB / L2T,
from which, by eliminating TB, we get:
|L1B1 · L1T / R1 = L2B2 · L2T,|
|L1B1 · (L1T / R1) = L2B2 · (L2T / R2).|
Now recollect that for point T on the circle of similitude the expressions in parentheses on both sides of the equality are equal, which implies
|L1B1 = L2B2,|
and subsequently (4).
The converse of the statement is proved in a standard way. Choose one of the circles, say, c1 and the corresponding L point, L1. Let the tangent to c1 at L1 intersect the circle of similitude in T. From T draw the tangents to c2 and select one of the points of contact as, say, L'2. (In order to select the right point of contact, observe that triangle L1TM1 and L2TM2 have different orientations.) By the direct statement,
|Pow(L'2, c1) = Pow(L1, c2) = Pow(L2, c1),|
from which, due to (5), L'2 = L2.
- J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971