Properties of the Circle of Similitude

Properties of the Circle of Similitude: What Is This about?
A Mathematical Droodle

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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.

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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). C_M and C_S are centers of segments O₁O₂ and PQ, respectively.

The circle of similitude (c_S) 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 C_S 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, c_S in particular, is orthogonal to the circle with O₁O₂ as a diameter. This is statement 1 hinted at by the applet.

Statement 1

The circle of similitude of any two circle with centers at O₁, O₂ is orthogonal to the circle with O₁O₂ as a diameter.

Statement 2

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 L_i, M_i lying on c_i, i = 1, 2, so that TP will serve as the bisector of angles L₁TL₂ and M₁TM₂:

(2)	∠L₁TP = ∠PTL₂.

Proof

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 C_S, 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.

Statement 3

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 H_i, i = 1, 2, is the projection of L_i on the radical axis of c₁ and c₂ then

(3)	L₁H₁ = L₂H₂.

Statement 4

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.

Statement 4'

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, t(L₁, c₂)² is the power of point L₁ with respect to circle c₂.

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.

t(L₁, c₂)²	= L₁K₂·L₁L₂
	= L₂K₁·L₂L₁
	= t(L₂, c₁)².

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. d_i, i = 1, 2, is the distance from S to O_i. We are interested in the difference of the powers of S with respect to two circles c₁, c₂:

Pow(S, c₁) - Pow(S, c₂)	= (d₁² - R₁²) - (d₂² - R₂²)
	= (d₁² - d₂²) - (R₁² - R₂²)
	= ((PH² + (FO₁ + FH)²) - ((PH² + (FO₂ - FH)²) - (R₁² - R₂²)
	= ((FO₁ + FH)² - (FO₂ - FH)²) - (R₁² - R₂²)
	= (FO₁² - R₁²) - (FO₂² - R₂²) + 2·(FO₁ + FO₂)·FH
	= 2·(FO₁ + FO₂)·FH,

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 B_i, i = 1, 2, be the midpoints of the segments L_iK_i. The configuration contains two pairs of similar right triangles: ΔL₁O₁B₁ is similar to ΔTL₁B₁, and ΔL₂O₂B₂ is similar to triangle ΔTL₂B₂. This is because, for example, angle K₁L₁T, formed by a tangent TL₁ and a secant BL₁ equals half the measure of the central angle L₁O₁K₁. The second circle is treated similarly.

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.

References

J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971

Radical Axis and Radical Center

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