Overview

Descartes' theorem is a formula in four variables which describes the relationship between the radii of 4 mutually tangent circles. It most commonly appears in AMC10/12 and some AIME problems.

Key Ideas

• The formal statement of the theorem is

$k_{1} + k_{2} + k_{3}+ k_{4} = \sqrt{2(k_{1} + k_{2} + k_{3} + k_{4})}$ where $k_{i} = \pm \frac{1}{r_{i}}$ and r is the length of the radii of the i-th circle. K is known as the "curvature" of the i-th circle.

• The sign of the curvature depends on whether or not the circle is internally or externally tangent to the other circles. For example,

in the worked example below, the curvatures are all positive because they are all externally tangent to eachother.

• Furthermore, Descartes' Theorem also works for lines instead of circles. For lines, the curvature is exactly 0, while the formula maintains its original form.

Worked Example

Three mutually tangent circles have radii $1$, $2$, and $3$. A fourth circle is tangent to all three. Find the radius of the smallest such circle.

We are given the radii of three circles and are asked to find an external tangent circle.

We use Descartes’ Theorem with curvatures (reciprocals of the radii): $k_1 = 1, \quad k_2 = \tfrac{1}{2}, \quad k_3 = \tfrac{1}{3}$

Descartes' Circle Formula states $(k_1 + k_2 + k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2)$

Plugging in: $\left(1 + \tfrac{1}{2} + \tfrac{1}{3} + k_4\right)^2 = 2\left(1 + \tfrac{1}{4} + \tfrac{1}{9} + k_4^2\right)$

Simplifying: $\left(\tfrac{11}{6} + k_4\right)^2 = 2\left(\tfrac{49}{36} + k_4^2\right)$

$\tfrac{121}{36} + \tfrac{11}{3}k_4 + k_4^2 = \tfrac{49}{18} + 2k_4^2$

$k_4^2 - \tfrac{11}{3}k_4 - \tfrac{23}{36} = 0$

Using the quadratic formula: $k_4 = \frac{\tfrac{11}{3} \pm \sqrt{\tfrac{121}{9} - (-\tfrac{23}{9})}}{2} = \frac{\tfrac{11}{3} \pm \sqrt{\tfrac{144}{9}}}{2} = \frac{\tfrac{11}{3} \pm \tfrac{12}{3}}{2} = \tfrac{23}{6}, -\tfrac{1}{6}$

Since curvature is the reciprocal of the radius, the two possible circles are: $r = \frac{1}{|k_4|} = \frac{6}{23}$ and $6$

The larger circle contains the three given circles, so the desired radius is $\frac{6}{23}$

More Examples

Strategy Checklist

• Look for keywords such as "mutually tangent" and "the smallest circle".
• If the problem involves 3 mutually tangent circles, see if Descartes' Theorem applies.
• keep in mind that the theorem also holds for lines aswell.

Common Pitfalls

• forgetting the sign of the curvatures.
• forgetting it also works for a line.
• incorrectly solving the quadratic.

Practice Problems