Properties of the Circle of Similitude: What Is This about?
A Mathematical Droodle
What if applet does not run? 
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Copyright © 19962018 Alexander BogomolnyThe 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 c_{1}, c_{2} are given by their centers O_{1}, O_{2} and radii R_{1}, R_{2}. The centers of similarity of c_{1} and c_{2} are denoted P (internal) and Q (external). C_{M} and C_{S} are centers of segments O_{1}O_{2} and PQ, respectively.
The circle of similitude (c_{S}) of c_{1} and c_{2} is an Apollonian circle with respect two the centers O_{1} and O_{2} and ratio R_{1}/R_{2} (or R_{2}/R_{1}). 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 O_{1}, O_{2}. Each member of the family, c_{S} in particular, is orthogonal to the circle with O_{1}O_{2} 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 O_{1}, O_{2} is orthogonal to the circle with O_{1}O_{2} as a diameter.
Statement 2
For a point T on the circle of similitude of two circles c_{1} and c_{2}, the four points of contact of the tangents from T to the circles can be split into two pairs L_{1}, L_{2} and M_{1}, M_{2}, with L_{i}, M_{i} lying on c_{i},
(2)  ∠L_{1}TP = ∠PTL_{2}. 
Proof
For a point T on the circle of similitude, the tangents two the two circles form similar triangles. This makes angles L_{1}TM_{1} and L_{2}TM_{2} equal. The lines TO_{1} and TO_{2} serve naturally is the bisectors of these angles. The points of tangency L_{1}, M_{1} for c1, and L_{2}, M_{2} for c2, can be paired  L_{1}, L_{2} and M_{1}, M_{2}  so that the angles PTL_{1} and PTL_{2} are equal as are the angles PTM_{1} and PTM_{2}. (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 O_{1}TO_{2}. 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 L_{1} and L_{2} are equidistant from the radical axis of c_{1} and c_{2}, and so are M_{1} and M_{2}. (This is the same as saying that the midpoint of L_{1}L_{2} (and also that of M_{1}M_{2}) lies on the radical axis.) If H_{i},
(3)  L_{1}H_{1} = L_{2}H_{2}. 
Statement 4
Let the line L_{1}L_{2} crosses c_{1} in K_{1} and c_{2} in K_{2}. Then
(4)  L_{1}K_{1} = L_{2}K_{2}. 
The converse is also true: If a line cuts two circles in four points K_{1}, L_{1}, L_{2}, K_{2} (in that order) so that (4) holds, then the tangents to the circles at L_{1} and L_{2} (as well as those at K_{1} and K_{2}) 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(L_{1}, c_{2}) = t(L_{2}, c_{1}). 
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 c_{1}, c_{2} with their line of centers. For a point S, let H be the projection of S on the latter. d_{i},

the latter because the powers of F with respect to the two circles c_{1}, c_{2} 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 O_{1}, O_{2}, we can write
Pow(S, c_{1})  Pow(S, c_{2}) = 2·d·FH. 
For a point on c_{2},
(5)  Pow(S, c_{1}) = 2·d·FH. 
A similar formula for a point on c_{1} 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 L_{1}L_{2}. (The following applies equally well to points M which, for this reason, won't be mentioned again.) Let B_{i},
From the aforementioned similarities we obtain the following proportions:
L_{1}B_{1} / R_{1} = TB / L_{1}T and L_{2}B_{2} / R_{2} = TB / L_{2}T, 
from which, by eliminating TB, we get:
L_{1}B_{1} · L_{1}T / R_{1} = L_{2}B_{2} · L_{2}T, 
or
L_{1}B_{1} · (L_{1}T / R_{1}) = L_{2}B_{2} · (L_{2}T / R_{2}). 
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
L_{1}B_{1} = L_{2}B_{2}, 
and subsequently (4).
The converse of the statement is proved in a standard way. Choose one of the circles, say, c_{1} and the corresponding L point, L_{1}. Let the tangent to c_{1} at L_{1} intersect the circle of similitude in T. From T draw the tangents to c_{2} and select one of the points of contact as, say, L'_{2}. (In order to select the right point of contact, observe that triangle L_{1}TM_{1} and L_{2}TM_{2} have different orientations.) By the direct statement,
Pow(L'_{2}, c_{1}) = Pow(L_{1}, c_{2}) = Pow(L_{2}, c_{1}), 
from which, due to (5), L'_{2} = L_{2}.
References
 J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS  Chelsea Publishing, 1971
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny