Descartes' Circle Theorem

Descartes' circle theorem (a.k.a. the kissing circle theorem) provides a quadratic equation satisfied by the radii of four mutually tangent circles. By solving this equation, one can determine the possible values for the radius of a fourth circle tangent to three given, mutually tangent circles. The theorem was first stated in a 1643 letter from René Descartes to Princess Elizabeth of the Palatinate, presumably in an attempt to impress her.

Descartes' circle theorem (a.k.a. the kissing circle theorem)

Suppose circles $C_i$ for $1\le i \le 4$ are mutually tangent, i.e., any two of the four circles touch at precisely one point. Let $r_i$ denote the radius of $C_i$.

The curvature of $C_i$ is defined to be $k_i = \pm \frac{1}{r_i},$ where the sign of $k_i$ is chosen depending on whether or not $C_i$ is internally or externally tangent to the other circles. For example, in the above picture, the large green circle has negative curvature, since the other three circles are internally tangent to it. On the other hand, the small red circle has positive curvature, since the other three circles are externally tangent to it.

Descartes' circle theorem states that the following quadratic equation holds:

$$(k_1 + k_2 + k_3 + k_4)^2 = 2\big(k_{1}^{2} + k_{2}^{2} + k_{3}^{2} + k_{4}^{2}\big).$$

In particular, if $k_1$, $k_2$, and $k_3$ are known, one can solve for $k_4$ as

$$k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1 k_2 + k_2 k_3 + k_1 k_3}.$$

These are the two explicit possibilities for $k_4$. One of these solutions is positive, and the other is either positive or negative; if the second solution is negative, it must represent a circle that is internally tangent to the other three (like the large green circle in the diagram above).

A straight line can be thought of as a circle with infinite radius, or equivalently, zero curvature. Thus, it is possible to set one of the curvatures in Descartes' equation equal to zero and obtain the relation describing a situation like the below picture:

The numbers displayed are the curvatures of the circles in the diagram.

Setting $k_3 = 0$, this relation is the simpler equation $k_4 = k_1 + k_2 \pm 2\sqrt{k_1 k_2}$. For example, using the numbers from the diagram above, one obtains $k_4 = 25 + 81 \pm 2\sqrt{25 \cdot 81} = 106 \pm 90 = \{196, 16\}$, as depicted.

Although Descartes' circle theorem is easily stated, it is not easy to prove. Below is a proof via trigonometry:

We start with the trigonometric identity that states if $\alpha+ \beta+ \gamma=2\pi$, then $${\cos}^{2}\alpha+{\cos}^{2}\beta+{\cos}^{2}\gamma=1+2 \cos\alpha \cos\beta \cos\gamma.$$ Proof of identity: Rewriting the above identity, we have $$2\left({\cos}^{2}\alpha+{\cos}^{2}\beta+{\cos}^{2}\gamma\right) -3=4\cos\alpha \cos\beta \cos\gamma -1.$$ We will show that $$LHS= RHS.$$ We start with $$\begin{aligned} S &= \cos(2\alpha ) + \cos(2\beta ) + \cos(2\gamma)\\ &=2\left({\cos}^{2}\alpha+{\cos}^{2}\beta+{\cos}^{2}\gamma\right) -3\\ &= LHS . \end{aligned}$$ Now, $$\begin{aligned} \cos(2\alpha)+\cos(2\beta) &=2\cos(\alpha + \beta )\cos(\alpha - \beta)\\ &=2\cos(2\pi - \gamma )\cos(\alpha - \beta)\\ &=2\cos(\gamma )\cos(\alpha - \beta). \end{aligned}$$ Then $$\begin{aligned} S &=2\cos(\gamma)\cos(\alpha - \beta)+2{\cos}^2(\gamma) - 1 \\ &=2\cos(\gamma)\big(\cos(\alpha - \beta)+\cos(\gamma)\big) - 1 \\ &=2\cos\big(\gamma)(\cos(\alpha - \beta)+\cos(\alpha + \beta)\big) - 1 \\ &= 4\cos\alpha \cos\beta \cos\gamma -1\\ &=RHS. \end{aligned}$$
This implies $LHS=RHS,$ i.e. $$2\left({\cos}^{2}\alpha+{\cos}^{2}\beta+{\cos}^{2}\gamma\right) -3=4\cos\alpha \cos\beta \cos\gamma -1.$$ Hence, we finally get our identity.

Helper diagram for proof

Now, triangle $ABC$ in the diagram to the right, where $A, B$ and $C$ are the centers of the three black circles, has side lengths $$\lvert\overline{AB}\rvert={r}_{1}+{r}_{2},\ \ \lvert\overline{BC}\rvert={r}_{2}+{r}_{3},\ \ \lvert\overline{CA}\rvert= {r}_{1}+{r}_{3}.$$ Let $O$ be the center of the small red circle with radius $r_4$ externally tangential to all of the three black circles, then $$\lvert\overline{AO}\rvert= {r}_{1}+{r}_{4},\ \ \lvert\overline{BO}\rvert={r}_{2}+{r}_{4},\ \ \lvert\overline{CO}\rvert={r}_{3}+{r}_{4}.$$ Also, let $$\angle AOB=\gamma,\ \ \angle BOC=\alpha,\ \ \angle AOC=\beta.$$ Then applying the cosine rule in triangles $AOB, BOC,$ and $AOC$ gives $$\begin{aligned} \cos(\gamma) &= \frac { { AO }^{ 2 }+{ BO }^{ 2 }-{ AB }^{ 2 } }{ 2 \cdot AO \cdot BO } \\ &=\frac { { ({r}_{4}+{ r }_{ 1 }) }^{ 2 }+{ ({r}_{4}+{ r }_{ 2 }) }^{ 2 }-{ ({ r }_{ 1 }+{ r }_{ 2 }) }^{ 2 } }{ 2({r}_{4}+{ r }_{ 1 })({r}_{4}+{ r }_{ 2 }) }\\ &=\frac { 2{ r }_{4}^{2}+2{r}_{4}({ r }_{ 1 }+{ r }_{ 2 })-2{ r }_{ 1 }{ r }_{ 2 } }{ 2({r}_{4}+{ r }_{ 1 })({r}_{4}+{ r }_{ 2 }) }\\ &=1-\frac { 2{ r }_{ 1 }{ r }_{ 2 } }{ ({r}_{4}+{ r }_{ 1 })({r}_{4}+{ r }_{ 2 }) } . \end{aligned}$$ Replacing radii with their respective curvatures, we get $$\cos(\gamma)=1-\frac { 2{ k }_{4}^{2} }{ ({k}_{4}+{ k }_{ 1 })({k}_{4}+{ k }_{ 2 }) }=1-{\lambda}_{3}.$$ Similarly, we have $$\begin{aligned} \cos(\beta)&=1-\frac { 2{ k }_{4}^{2} }{ ({k}_{4}+{ k }_{ 1 })({k}_{4}+{ k }_{ 3 }) }=1-{\lambda}_{2}\\ \cos(\alpha)&=1-\frac { 2{ k }_{4}^{2} }{ ({k}_{4}+{ k }_{ 2 })({k}_{4}+{ k }_{ 3 }) }=1-{\lambda}_{1} . \end{aligned}$$ Placing the values of $\cos(\alpha), \cos(\beta), \cos(\gamma)$ on our trignometric identity, we get $${ (1-{ \lambda }_{ 1 }) }^{ 2 }+{ (1-{ \lambda }_{ 2 }) }^{ 2 }+{ (1-{ \lambda }_{ 3 }) }^{ 2 }=1+2(1-{ \lambda }_{ 1 })(1-{ \lambda }_{ 2 })(1-{ \lambda }_{ 3 }).$$ Simplifying, we get $${ { \lambda }_{ 1 } }^{ 2 }+{ { \lambda }_{ 2 } }^{ 2 }+{ { \lambda }_{ 3 } }^{ 2 }+2{ \lambda }_{ 1 }{ \lambda }_{ 2 }{ \lambda }_{ 3 }=2({ \lambda }_{ 1 }{ \lambda }_{ 2 }+{ \lambda }_{ 2 }{ \lambda }_{ 3 }+{ \lambda }_{ 1 }{ \lambda }_{ 3 }).$$ Dividing both sides by ${ \lambda }_{ 1 }{ \lambda }_{ 2 }{ \lambda }_{ 3 }$ gives $$\frac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 }{ \lambda }_{ 3 } } +\frac { { \lambda }_{ 2 } }{ { \lambda }_{ 1 }{ \lambda }_{ 3 } } +\frac { { \lambda }_{ 3 } }{ { \lambda }_{ 1 }{ \lambda }_{ 2 } } +2=2\left(\frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } +\frac { 1 }{ { \lambda }_{ 3 } } \right).$$ Putting the values of ${\lambda}_{1},{\lambda}_{2},{\lambda}_{3}$ gives $$\frac { { ({k}_{4}+{ k }_{ 1 }) }^{ 2 } }{ 2{ k }_{4}^{2} } +\frac { { ({k}_{4}+{ k }_{ 2 }) }^{ 2 } }{ 2{ k }_{4}^{2} } +\frac { { ({k}_{4}+{ k }_{ 3 }) }^{ 2 } }{ 2{ k }_{4}^{2} } +2=2\left(\frac { ({k}_{4}+{ k }_{ 2 })({k}_{4}+{ k }_{ 3 }) }{ 2{ k }_{4}^{2} } +\frac { ({k}_{4}+{ k }_{ 1 })({k}_{4}+{ k }_{ 2 }) }{ 2{k}_{4}^{2} } +\frac { ({k}_{4}+{ k }_{ 1 })({k}_{4}+{ k }_{ 2 }) }{ 2{ k }_{4}^{2} } \right).$$ Simplifying, we get $${ k }_{ 1 }^{ 2 }+{ k }_{ 2 }^{ 2 }+{ k }_{ 3 }^{ 2 }+2k_4({ k }_{ 1 }+{ k }_{ 2 }+{ k }_{ 3 })+7{ k }_{4}^{2}=6{ k }_{4}^{2}+4k_4({ k }_{ 1 }+{ k }_{ 2 }+{ k }_{ 3 })+2({ k }_{ 1 }{ k }_{ 2 }+{ k }_{ 2 }{ k }_{ 3 }+{ k }_{ 1 }{ k }_{ 3 }).$$ Further simplifying, we get $$\begin{aligned} { k }_{ 1 }^{ 2 }+{ k }_{ 2 }^{ 2 }+{ k }_{ 3 }^{ 2 }+{ k }_{ 4 }^{ 2 } &=2{ k }_{ 4 }({ k }_{ 1 }+{ k }_{ 2 }+{ k }_{ 3 })+2({ k }_{ 1 }{ k }_{ 2 }+{ k }_{ 2 }{ k }_{ 3 }+{ k }_{ 1 }{ k }_{ 3 })\\ &={({k}_{1}+{k}_{2}+{k}_{3}+{k}_{4})}^{2}-\left({k}_{1}^{2}+{k}_{2}^{2}+{k}_{3}^{2}+{k}_{4}^{2}\right). \end{aligned}$$ Finally, we have $$2\left({ k }_{ 1 }^{ 2 }+{ k }_{ 2 }^{ 2 }+{ k }_{ 3 }^{ 2 }+{ k }_{ 4 }^{ 2 }\right)={ ({ k }_{ 1 }+{ k }_{ 2 }+{ k }_{ 3 }+{ k }_{ 4 }) }^{ 2 }.$$ Hence, the theorem is proved. $_\square$

Descartes' circle theorem can also be generalized to non-2D space, in a theorem sometimes referred to as the Soddy-Gosset theorem. In Euclidean space of $n$ dimensions, the maximum number of mutually tangent spheres is $n+2$. In 2-dimensional space, that's 4 mutually tangent circles; in 3-dimensional space, 5 mutually tangent spheres can always be found. The relationship of their curvatures is

$$n \left( \sum_{i=1}^{n+2} \kappa_i^2 \right) = \left(\sum_{i=1}^{n+2}\kappa_i\right)^2.$$