# 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.

#### Contents

## Statement of Theorem

Suppose circles \(C_i\), \(1\le i \le 4\) 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).

## Special Case of Degenerate Circle

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:

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.

## Proof Using Trigonometry

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 \[\begin{align} S &= \cos(2\alpha ) + \cos(2\beta ) + \cos(2\gamma)\\ &=2\left({\cos}^{2}\alpha+{\cos}^{2}\beta+{\cos}^{2}\gamma\right) -3\\ &= LHS . \end{align}\] Now, \[\begin{align} \cos(2\alpha)+\cos(2\beta) &=2\cos(\alpha + \beta )\cos(\alpha - \beta)\\ &=2\cos(2\pi - \gamma )\cos(\alpha - \beta)\\ &=2\cos(\gamma )\cos(\alpha - \beta). \end{align}\] Then \[\begin{align} S &=2\cos(\gamma)\cos(\alpha - \beta)+\big(2{\cos}(\gamma)\big)^2 - 1 \\ &=2\cos(\gamma)\big(\cos(\alpha - \beta)+\cos(\gamma)\big) - 1 \\ &=2\cos\big(\gamma)(\cos(\alpha - \beta)+\cos(\alpha + \beta)\big) - 1 \\ &= 4\cos\alpha \cos\beta \cos\gamma -1\\ &=RHS. \end{align} \]

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.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 \[\begin{align} \cos(\gamma) &= \frac { { AO }^{ 2 }+{ BO }^{ 2 }-{ AB }^{ 2 } }{ 2.AO.BO } \\ &=\frac { { ({r}_{4}+{ r }_{ 1 }) }^{ 2 }+{ ({r}_{4}+{ r }_{ 2 }) }^{ 2 }-{ ({ r }_{ 1 }+{ r }_{ 2 }) }^{ 2 } }{ 2({r}_{4}+{ r }_{ 1 })({r}_{4}+{ r }_{ 2 }) }\\ &=\frac { 2{ r }_{4}^{2}+2{r}_{4}({ r }_{ 1 }+{ r }_{ 2 })-2{ r }_{ 1 }{ r }_{ 2 } }{ 2({r}_{4}+{ r }_{ 1 })({r}_{4}+{ r }_{ 2 }) }\\ &=1-\frac { 2{ r }_{ 1 }{ r }_{ 2 } }{ ({r}_{4}+{ r }_{ 1 })({r}_{4}+{ r }_{ 2 }) } . \end{align}\] 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 \[\begin{align} \cos(\beta)&=1-\frac { 2{ k }_{4}^{2} }{ ({k}_{4}+{ k }_{ 1 })({k}_{4}+{ k }_{ 3 }) }=1-{\lambda}_{2}\\ \cos(\alpha)&=1-\frac { 2{ k }_{4}^{2} }{ ({k}_{4}+{ k }_{ 2 })({k}_{4}+{ k }_{ 3 }) }=1-{\lambda}_{1} . \end{align}\] 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 \[\begin{align} { k }_{ 1 }^{ 2 }+{ k }_{ 2 }^{ 2 }+{ k }_{ 3 }^{ 2 }+{ k }_{ 4 }^{ 2 } &=2{ k }_{ 4 }({ k }_{ 1 }+{ k }_{ 2 }+{ k }_{ 3 })+2({ k }_{ 1 }{ k }_{ 2 }+{ k }_{ 2 }{ k }_{ 3 }+{ k }_{ 1 }{ k }_{ 3 })\\ &={({k}_{1}+{k}_{2}+{k}_{3}+{k}_{4})}^{2}-\left({k}_{1}^{2}+{k}_{2}^{2}+{k}_{3}^{2}+{k}_{4}^{2}\right). \end{align}\] 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. \(_\square\)

## Generalization to \(n\) Dimensions

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

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

**Cite as:**Descartes' Circle Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/descartes-theorem/