Kissing Circles (Three Circles and a Line)

Circles that are mutually tangent to each other are called “kissing circles” because they barely touch each other (or “kiss”) at one point.  There are several different diagrams that have kissing circles, but this article will focus on the specific case of three circles and a line that are all mutually tangent to each other, as in the diagram below:

When three circles and a line are mutually tangent to each other, the relationship between the three radii of each circle is

The proof can be found by solving the three right triangles formed by using the segments from one center of a circle to the next as hypotenuses, as pictured below in gray:

The bottom left gray triangle makes the Pythagorean equation (r₁ – r₃)² + x₁² = (r₁ + r₃)², which simplifies to

(r1 – r3)2 + x12 = (r1 + r3)2	(Pythagorean’s Theorem)
r12 – 2r1r3 + r32 + x12 = r12 + 2r1r3 + r32	(expand)
-2r1r3 + x12 = 2r1r3	(subtract r12 and r32)
x12 = 4r1r3	(add 2r1r3)
x1 = 2√(r1r3)	(square root)

The bottom right gray triangle makes the Pythagorean equation (r₂ – r₃)² + x₂² = (r₂ + r₃)², which similarly simplifies to x₂ = 2√(r₂r₃), and the top gray triangle makes (r₁ – r₂)² + x₃² = (r₁ + r₂)², which simplifies to x₃ = 2√(r₁r₂).  Since x₁ + x₂ = x₃,

x1 + x2 = x3	(angle addition)
2√(r1r3) + 2√(r2r3) = 2√(r1r2)	(substitution)
√(r1r3) + √(r2r3) = √(r1r2)	(divide by 2)
√r3(√r1 + √r2) = √(r1r2)	(factor √r3)
√r3 = √(r1r2)/(√r1 + √r2)	(divide by (√r1 + √r2))
1/√r3 = (√r1 + √r2) /√(r1r2)	(rearrange)
1/√r3 = √r1/√(r1r2) + √r2/√(r1r2)	(distribute denominator)
1/√r3 = 1/√r2 + 1/√r1	(simplify)
1/√r1 + 1/√r2 = 1/√r3	(rearrange)

The mathematician Descartes studied the case of four kissing circles and derived the equation

where k₁ = ¹/_r1, k₂ = ¹/_r2, k₃ = ¹/_r3, and k₄ = ¹/_r4.

Four Kissing Circles

The derivation for this formula involves solving a quadratic system of three equations of three variables and is beyond the scope of this article.  However, we can use Descartes’ Theorem to as another proof for ¹/_√r1 + ¹/_√r2 = ¹/_√r3, because the case of three kissing circles and a line is actually a specific case in which the line has a curvature of k₃ = 0.  Therefore,

k4 = k1 + k2 + k3 ± 2√(k1k2 + k2k3 + k1k3)	(Descartes’ Theorem)
k4 = k1 + k2 + k3 + 2√(k1k2 + k2k3 + k1k3)	(equation for inner circle)
k4 = k1 + k2 + (0) + 2√(k1k2 + k2(0) + k1(0))	(k3 = 0)
k4 = k1 + k2 + 2√(k1k2)	(simplify)
k4 = k1 + 2√(k1k2) + k2	(rearrange)
k4 = (√k1 + √k2)2	(factor)
√k4 = √k1 + √k2	(square root)
1/√r4 = 1/√r2 + 1/√r1	(substitute)
1/√r1 + 1/√r2 = 1/√r4	(rearrange)

The relationship between the three radii can also be written as r₁^-½ + r₂^-½ = r₃^-½, or a^-½ + b^-½ = c^-½, which is fascinatingly similar to relationship of the three sides in a right angle, which according to Pythagorean’s Theorem is a² + b² = c².  The only difference between the two equations is the exponents, which are opposite reciprocals.

Now that we have established the relationship of the radii for three kissing circles and a line, we can investigate some specialized cases that produce some unexpected sequences.  It should be noted that if two radii are the reciprocal of a square integer, then the third radius will be the reciprocal of the sum of those two integers squared.  In other words, if r₁ = ¹/_m2 and r₂ = ¹/_n2, then r₃ = ¹/_{(m + n)}2.  This can be shown algebraically:

1/√r1 + 1/√r2 = 1/√r3	(radius relationship)
r1-½ + r2-½ = r3-½	(rewrite exponents)
(1/m2)-½ + (1/n2)-½ = r3-½	(substitute  r1 and r2)
m + n = r3-½	(simplify exponents)
(m + n)-2 = r3	(negative square both sides)
r3 = 1/(m + n)2	(rewrite exponent)

So if two kissing circles have a radii of 1, then r₁ = 1 = ¹/₁2 and m = 1 and r₂ = 1 = ¹/₁2 and n = 1, and the third kissing circle that is in between them will have a radius r₃ = ¹/_{(m + n)}2 = ¹/_{(1 + 1)}2 = ¹/₂2.  Continuing on in one direction, the circle between this new circle with radius ¹/₂2 and the original circle ¹/₁2 will have a radius of ¹/_{(2 + 1)}2 = ¹/₃2, and the circle between this new circle with radius ¹/₃2 and the original circle ¹/₁2 will have a radius of ¹/_{(3 + 1)}2 = ¹/₄2, and so on, so that sequence of new radii for each successive circle is r = ¹/₁2, ¹/₂2, ¹/₃2, ¹/₄2, … which are the square reciprocals of the counting numbers.

Square Reciprocals of Counting Numbers

Similarly, if we start with two kissing circles that have a radii of 1 but alternate directions left and right for every other circle, the sequence becomes r = ¹/₁2, ¹/₁2, ¹/_{(1 + 1)}2 =¹/₂2, ¹/_{(1 + 2)}2 =¹/₃2, ¹/_{(2 + 3)}2 =¹/₅2, ¹/_{(3 + 5)}2 =¹/₈2, … or r = ¹/₁2, ¹/₁2, ¹/₂2, ¹/₃2, ¹/₅2, ¹/₈2, … which are the square reciprocals of the Fibonacci numbers.

                                                                                                                                                  

Square Reciprocals of Fibonacci Numbers

One last specialized case of kissing circles and a line that should be mentioned are Ford circles, named after the twentieth century mathematician Lester R. Ford.  Starting with two circles that have radii of r = ½, place the center of one circle directly above the 0 on the number line and place the center of the other circle directly above the 1.  The new kissing circle formed in between these two circles will have a center directly above x = ¹/₂.  Furthermore, the circle between the circles x = 0 and x = ¹/₂ is x = ¹/₃, and the circle between the circles x = ¹/₂ and x = 1 is x = ²/₃.  This process can be continued on indefinitely, where it is possible to place a new circle at the x-value of any irreducible fraction ^p/_q.

                        

Ford Circles

Additionally, it can be proved that the kissing circle between any x = ^p1/_q1 and x = ^p2/_q2 is x = ^{(p1 + p2)}/ _{(q1 + q2)}.  For example, the kissing circle between the circles x = ¹/₂ and x = ²/₃ is x = ^{(1 + 2)}/_{(2 + 3)} = ³/₅. 

                                      

To prove this, we must first recognize that the first two outer Ford circles both have x-values in the form of x = ^p/_q and radii in the form of r = ¹/_2q2.  (For the left circle, p₁ = 0 and q₁ = 1, so that x = ^p1/_q1 = 0 and r = ¹/_2q12 = ¹/₂.  For the right circle, p₂ = 1 and q₂ = 1, so that x = ^p2/_q2 = 1 and r = ¹/_2q22 = ¹/₂.)  It can then be proved algebraically that any new kissing circle produced by two circles with both x-values in the form of x = ^p/_q and radii in the form of r = ¹/_2q2 also have these properties, specifically x = ^{(p1 + p2)}/ _{(q1 + q2)} and r = ¹/_{2(q1 +q2)}2.  First, to prove that r = ¹/_{2(q1 +q2)}2:

1/√r1 + 1/√r2 = 1/√r3	(radius relationship)
r1-½ + r2-½ = r3-½	(rewrite exponents)
(1/2q12)-½ + (1/2q22)-½ = r3-½	(subst r1 and r2)
2½q1 + 2½q2 = r3-½	(simplify exponents)
2½(q1 + q2) = r3-½	(factor)
2-1(q1 + q2)-2 = r3	(negative square both sides)
r3 = 1/2(q1 +q2)2	(rewrite exponent)

To prove that x = ^{(p1 + p2)}/ _{(q1 + q2)}, recall from above that x₃ is the distance between the centers of the two outside kissing circles, which in this case is x₃ = ^p2/_q2 – ^p1/_q1, and that x₁ is the distance between the centers of an outside and inside kissing circle, which in this case we are trying to prove that x₁ = ^{(p1 + p2)}/ _{(q1 + q2)} – ^p1/_q1.  Also recall from above that x₁ = 2√(r₁r₃), x₃ = 2√(r₁r₂), and √r₃ = ^√(r1r2)/_{(√r1 + √r2)}.  Therefore:

x1 = 2√(r1r3)	(given)
x1 = 2√r1√r3	(distribute)
x1 = 2√r1√(r1r2)/(√r1 + √r2)	(subst √r3 = √(r1r2)/(√r1 + √r2))
x1 = √r1x3/(√r1 + √r2)	(subst x3 = 2√(r1r2))
x1 = √(1/2q12)(p2/q2 – p1/q1) / (√(1/2q12) + √(1/2q22))	(subst r1 = 1/2q12, r2 = 1/2q22, and x3 = p2/q2 – p1/q1)
x1 = 1/q1(p2/q2 – p1/q1) / (1/q1 + 1/q2)	(simplify)
x1 = (q1p2 – q2p1)/q1(q2 + q1)	(multiply by q1q1q2/q1q1q2)
x1 = (q1p2 + q1p1 – q1p1 – q2p1)/q1(q2 + q1)	(add and subtract q1p1)
x1 = q1(p2 + p1) – p1(q1 – q2)/q1(q2 + q1)	(factor)
x1 = q1(p2 + p1)/q1(q2 + q1) – p1(q1 – q2)/q1(q2 + q1)	(distribute denominator)
x1 = (p2 + p1)/(q2 + q1) – p1/q1	(simplify)

In conclusion, the relationship between the three kissing circles and a line is ¹/_√r1 + ¹/_√r2 = ¹/_√r3.  This can be rewritten as r₁^-½ + r₂^-½ = r₃^-½, which is fascinatingly similar to Pythagorean’s Theorem.  Furthermore, it can be shown that the radii of a series of kissing circles under two kissing unit circles and a line in one direction is a sequence of square reciprocals of the counting numbers, and in alternating directions is a sequence of square reciprocals of the Fibonacci numbers.  Finally, Ford circles are kissing circles in which it is possible to place a new circle at the x-value of any irreducible fraction ^p/_q, and any successive kissing circle between x = ^p1/_q1 and x = ^p2/_q2 will have a new x-value of x = ^{(p1 + p2)}/ _{(q1 + q2)}.