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Descartes Circle Theorem — Definition, Formula & Examples

The Descartes Circle Theorem is a formula that connects the curvatures (reciprocals of radii) of four circles that are all tangent to each other. Given three mutually tangent circles, it lets you find the radius of a fourth circle that fits snugly in the gap between them.

If four mutually tangent circles have curvatures k1,k2,k3,k4k_1, k_2, k_3, k_4 where ki=1rik_i = \frac{1}{r_i}, then (k1+k2+k3+k4)2=2(k12+k22+k32+k42)(k_1 + k_2 + k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2). A circle that encloses the others is assigned a negative curvature.

Key Formula

(k1+k2+k3+k4)2=2(k12+k22+k32+k42)(k_1 + k_2 + k_3 + k_4)^2 = 2\left(k_1^2 + k_2^2 + k_3^2 + k_4^2\right)
Where:
  • kik_i = Curvature of the $i$-th circle, equal to $\frac{1}{r_i}$ (negative if the circle encloses the others)
  • rir_i = Radius of the $i$-th circle

How It Works

Each circle's curvature is defined as k=1rk = \frac{1}{r}, where rr is the radius. If a large circle surrounds the other three, its curvature is taken as negative. You plug the three known curvatures into the formula and solve for the fourth. The formula yields a quadratic in k4k_4, which usually gives two solutions corresponding to the two different circles tangent to all three (one fitting in the small gap, one being the outer enclosing circle).

Worked Example

Problem: Three mutually tangent circles have radii r1=1r_1 = 1, r2=1r_2 = 1, and r3=1r_3 = 1. Find the radius of the small circle that fits in the gap between them.
Find curvatures: Since all three radii equal 1, each curvature is k=1/1=1k = 1/1 = 1.
k1=k2=k3=1k_1 = k_2 = k_3 = 1
Apply the theorem: Substitute into the Descartes formula and solve for k4k_4.
(1+1+1+k4)2=2(1+1+1+k42)(1 + 1 + 1 + k_4)^2 = 2(1 + 1 + 1 + k_4^2)
Solve: Expand the left side: (3+k4)2=9+6k4+k42(3 + k_4)^2 = 9 + 6k_4 + k_4^2. Set equal to 2(3+k42)=6+2k422(3 + k_4^2) = 6 + 2k_4^2. This gives 9+6k4+k42=6+2k429 + 6k_4 + k_4^2 = 6 + 2k_4^2, so k42−6k4−3=0k_4^2 - 6k_4 - 3 = 0. Using the quadratic formula:
k4=6±36+122=3±23k_4 = \frac{6 \pm \sqrt{36 + 12}}{2} = 3 \pm 2\sqrt{3}
Interpret: The positive solution k4=3+23≈6.464k_4 = 3 + 2\sqrt{3} \approx 6.464 gives the small inner circle. Its radius is r4=1/k4r_4 = 1/k_4.
r4=13+23=3−239−12=23−33≈0.155r_4 = \frac{1}{3 + 2\sqrt{3}} = \frac{3 - 2\sqrt{3}}{9 - 12} = \frac{2\sqrt{3} - 3}{3} \approx 0.155
Answer: The small circle that fits in the gap has radius r4=23−33≈0.155r_4 = \frac{2\sqrt{3} - 3}{3} \approx 0.155.

Why It Matters

The Descartes Circle Theorem appears in fractal geometry when building Apollonian gaskets — infinitely nested circle packings. It also connects to number theory, since when the initial curvatures are all integers, every subsequent curvature in the packing is also an integer.

Common Mistakes

Mistake: Forgetting to assign a negative curvature to an enclosing circle.
Correction: When one circle contains the other three, its curvature enters the formula as k=−1/rk = -1/r. Without the negative sign, the formula gives incorrect results.