If we are given two tangent circles how do we find a third circle tangent to both of them? If the two circles are internally tangent, the third circle is also internally tangent to the larger circle.

If the two circles are externally tangent, the third circle is either internally or externally tangent to both.

Note that we disregard cases where the third circle has the same point of tangency as the other two such as shown below.

**Locus of centre of third circle**

If we let the centres of the three circles be and with respective radii (assume ) then what is the locus of as are fixed while is allowed to vary?

We note in the internally tangent case,

Since this distance is independent of , is on an ellipse with foci at and shown in green below.

In the externally tangent case, there are two subcases depending on whether the third circle is internally or externally tangent to the other two:

This corresponds to the two branches of a hyperbola with foci at and shown in green below.

Notice that similar results hold when the original two circles are not necessarily tangent to each other.

**Distances**

Denoting the points of tangency between by and similarly defining , we shall find some distances in the figure below.

By the cosine rule in ,

The perimeter of is , so by Heron’s formula its area is

If the points have coordinates respectively, then has coordinates . We have found above and can be found via

Hence the coordinates of are

Note here that the first coordinate varies with linearly for fixed. The distance is found from

Hence we have found that interestingly the power of the tangency point with respect to circle is the relatively simple form .

We may also find the distance between points of tangency as using the identity . From (4),

Next we perform the same calculations (4)-(9) with the configuration of three circles tangent externally.

By the cosine rule in ,

The perimeter of is , so by Heron’s formula its area is the attractive form

If the points have coordinates respectively, then has coordinates . We have found above and can be found via

Hence the coordinates of are

Note again that the first coordinate varies with linearly for fixed. Formulas (7) and (13) were used to generate the diagrams in this post.

The distance is found from

Hence we have found that the power of the tangency point with respect to circle is .

The distance between points of tangency is and using the identity with (10),

Many of the formulas here are found from the first case by replacing with . Other distances between points in the figure can be found through similar means.

Finally, we remark that the radical axes of each pair of circles intersect at the radical centre – that point is the circumcentre of the triangle passing through the three points of tangency. With this circle being orthogonal to the three existing circles, inversion in this circle preserves the figure.

**Four circles**

For the case of four mutually tangential circles (where there is more than one point of tangency), there is an amazing relationship involving the complex number coordinates of the centres of the circles and their curvatures (reciprocals of radii) :

This is known as the complex Descartes theorem. Note that for this formula to be valid for internally tangent circles, we make the sign of the curvature of the larger circle negative.

Interestingly this was only proven in 2001 in [1] with a nice proof shown in [2]. That proof assigns a sphere to each of the six points of tangency with curvature equal to the sums of the curvatures of the two circles meeting there. It then turns out that three spheres corresponding to the tangency points on the same circle are mutually tangent. If one does the same construction of spheres for the dual configuration of circles shown in blue below (the original four circles are in black), the same set of six spheres is obtained!

Note that we have more results from (16). Since the relationships remain when the points are translated by the same complex number we have

Expanding this as a quadratic in and comparing coefficients of [1] gives further two relationships:

Equation (18) is known as Descartes’ Theorem, discovered in 1643.

To see (16) and (18) in action, let us set for example . Then applying (7) we find

From (18) we may write

which for (negative as the larger circle is internally tangent), leads to [ or .]

From (16) we may write

which for leads to

This is used to generate the sketch below. As can be seen there are two possible circles shown in red.

Here is how one may prove (19) without the insight of the dual configuration, alluded to in [3] and [4]. Let us consider the externally tangent case with two possible fourth circles shown in red.

If we perform an inversion of the configuration in a circle of radius centred at the tangency point of and , the circles and invert to parallel lines while the circles invert to equal-radii circles tangent to the parallel lines as shown below. Note that in this new configuration it is easier to see that there are two possible choices for .

What distance separates the two parallel lines? We need only look at the image of the points diametrically opposite in and . This leads to the distance . Hence the radii of the images of and is half this, or .

If we know the distance of the centre of the image of from , the following lemma will enable us to determine . We shall use the following useful lemma which tells us how to find the radius of a circle from quantities involving its inverse.

**Lemma**

If a circle has radius and distance from the origin, its inverse in a circle of radius centred at the origin has radius and its centre has distance from the origin. (Hence .)

**Proof of lemma**

In the above figure the dashed circle is the circle of inversion while the separate cases and are shown in blue and red respectively.

The points on the original circle collinear with the origin are at distance and from the origin. The inverses of these points have distance and from the origin (this quantity is negative if ). Hence the centre of the inverted circle has distance from the origin and radius as required. Note that if the circle inverts to a line which can be thought of as a circle with centre at infinity with infinite radius.

From (12), if the origin is at , has coordinates

Suppose has coordinates (we note that while is the image of under the inversion, the centre is generally not the image of ). Then is at distance

above or below (the distance between the parallel lines), so has coordinates . By the above lemma with , the radius of with centre is . Hence its curvature is

Firstly from (13)

, so by the lemma, we find via the ratio of radii to :

so

Next,

Using (28) and (29) in (26) gives us

With , this becomes establishing (21).

For the case of and internally tangent to a similar computation to the above can be carried out and we obtain

which is the same as (28) with replaced with .

Note that Descartes’ theorem also holds in the special cases of two parallel lines tangent to two equal circles or a single line tangent to three mutually tangent circles. In the latter case we may set and assuming we obtain from (18) the nice formula

We see these two solutions below in the right figure.

Further generalisations of Descartes’ theorem to further dimensions are in [1], and more can be read about tangent circle packings in [5].

#### References

[1] J.C. Lagarias, C.L. Mallows, and A. Wilks, * Beyond the Descartes Circle Theorem*, American Mathematical Monthly, **109** (2002), 338-361. Available at http://www.arxiv.org/abs/math?papernum=0101066.

[2] S. Northshield, Complex Descartes Circle Theorem, to appear, American Mathematical Monthly.

[3] D. Pedoe, On a theorem in geometry, American Mathematical Monthly, **74** (1967), 627-640.

[4] P. Sarnak, Integral Apollonian Packings, American Mathematical Monthly, **118** (2011), 291-306.

[5] D. Austin, When Kissing Involves Trigonometry, Feature Column from the AMS, March 2006.

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