## 1. Circle of similitude of two circles

Given two circles {c(O,r),c'(O',r')} and a point A (relative positions as in the figure below), construct the triangle of tangents ABC. This is similar to OO'A if and only if BC passes through the projection D of A on OO'.

If triangles {ABC,OO'A} are similar, then CO'DA is cyclic, D being the intersection point of {OO',BC}.
But O'CA is a right angle, hence O'DA is a right angle too.
Inversely assuming the above configuration, CO'DA is again cyclic. This time because at C, D are right angles. This implies that the angles at O', C are equal.
But ABOD is then also cylic, hence angle BAO is equal to BDO which is equal to CAO'. This implies that the angles at A of ABO' and AOO' are equal.

Remark-1 This restricts points A which may serve as similarity centers for similarities mapping (c) to (c'). They have to lie on an Apollonian circle (d) for the segment OO'. In fact, if A is a similarity center for the two circles {c,c'}, then triangles ABO, ACO' being similar implies AO/AO' = r/r'. Thus, A is on an Apollonian circle.
Remark-2 The two homothety centers of the circles H1, H2 are diametral points of circle (d).
Remark-3 Analogous properties hold in the case of other configurations for which the two circles are not external to each other.
The circle on diameter H1H2 is called the circle of similitude of the two circles.

## 2. Necessary and sufficient condition

A point A belongs to the circle of similitude of two circles if and only if, drawing two tangents {AC,AD} from A not separating the centers, the line of contacts cuts from the circles equal segments (DD'=CC').

The equality of the two segments is equivalent to the equality of powers of {C,D} with respect to the circles {k2, k1} correspondingly. The necessity is implied by an easy calculation. If s=r2/r1 denotes the similarity-ratio, then:

This implies that the projections {B',E'} of {B,E} on {AD,AC} respectively, together with {D,C} build a cyclic quadrilateral. Hence triangles CAB' and DAE' are similar and by the cyclic quadrangles {DAE'E, BB'AC} also the right-angled triangles AEE' and ABB' are similar. This implies that AE/AB = r2/r1 therefore proving that A is on the similitude circle of {k1, k2}.

## 3. Viewing under equal angles

The circle of similitude of two circles lying externally to each other is the locus of points A viewing the two circles under equal angles (i.e. the tangents to the two circles make equal angles at A).

This follows trivially by drawing lines {AE, AB} and calculating their ratio in terms of the radii of the two circles.

Remark The properties listed here are of importance in proving the theorem of Malfatti, after an idea of Steiner, who left his argument unproven until Hart completed it. An exposition of this story (following closely Coolidge) can be found in Malfatti.html ).