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
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(Pythagorean’s Theorem)
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r12 – 2r1r3 + r32
+ x12 = r12 + 2r1r3
+ r32
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(expand)
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-2r1r3 + x12 = 2r1r3
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(subtract r12 and r32)
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x12 = 4r1r3
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(add 2r1r3)
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x1 = 2√(r1r3)
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(square root)
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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
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(angle addition)
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2√(r1r3) + 2√(r2r3) =
2√(r1r2)
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(substitution)
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√(r1r3) + √(r2r3) = √(r1r2)
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(divide by 2)
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√r3(√r1 + √r2) = √(r1r2)
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(factor √r3)
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√r3 = √(r1r2)/(√r1 + √r2)
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(divide by (√r1 + √r2))
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1/√r3 = (√r1 + √r2) /√(r1r2)
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(rearrange)
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1/√r3 = √r1/√(r1r2) +
√r2/√(r1r2)
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(distribute denominator)
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1/√r3 = 1/√r2 +
1/√r1
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(simplify)
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1/√r1 + 1/√r2 =
1/√r3
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(rearrange)
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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)
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(Descartes’ Theorem)
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k4 = k1 + k2 + k3 + 2√(k1k2
+ k2k3 + k1k3)
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(equation for inner circle)
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k4 = k1 + k2 + (0) + 2√(k1k2
+ k2(0) + k1(0))
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(k3 = 0)
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k4 = k1 + k2 + 2√(k1k2)
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(simplify)
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k4 = k1 + 2√(k1k2) + k2
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(rearrange)
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k4 = (√k1 + √k2)2
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(factor)
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√k4 = √k1 + √k2
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(square root)
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1/√r4 = 1/√r2 +
1/√r1
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(substitute)
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1/√r1 + 1/√r2 =
1/√r4
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(rearrange)
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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
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(radius relationship)
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r1-½ + r2-½ = r3-½
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(rewrite exponents)
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(1/m2)-½ + (1/n2)-½
= r3-½
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(substitute r1 and
r2)
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m + n = r3-½
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(simplify exponents)
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(m + n)-2 = r3
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(negative square both sides)
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r3 = 1/(m + n)2
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(rewrite exponent)
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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
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(radius relationship)
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r1-½ + r2-½ = r3-½
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(rewrite exponents)
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(1/2q12)-½ + (1/2q22)-½
= r3-½
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(subst r1 and r2)
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2½q1 + 2½q2 = r3-½
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(simplify exponents)
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2½(q1 + q2) = r3-½
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(factor)
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2-1(q1 + q2)-2 = r3
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(negative square both sides)
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r3 = 1/2(q1 +q2)2
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(rewrite exponent)
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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)
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(given)
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x1 = 2√r1√r3
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(distribute)
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x1 = 2√r1√(r1r2)/(√r1 + √r2)
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(subst √r3 = √(r1r2)/(√r1 + √r2))
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x1 = √r1x3/(√r1 + √r2)
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(subst x3 = 2√(r1r2))
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x1 = √(1/2q12)(p2/q2
– p1/q1) /
(√(1/2q12) + √(1/2q22))
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(subst r1 = 1/2q12, r2 = 1/2q22,
and x3 = p2/q2 – p1/q1)
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x1 = 1/q1(p2/q2
– p1/q1) /
(1/q1 + 1/q2)
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(simplify)
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x1 = (q1p2 – q2p1)/q1(q2 + q1)
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(multiply by q1q1q2/q1q1q2)
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x1 = (q1p2 + q1p1 – q1p1 – q2p1)/q1(q2 + q1)
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(add and subtract q1p1)
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x1 = q1(p2 + p1) – p1(q1 – q2)/q1(q2 + q1)
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(factor)
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x1 = q1(p2 + p1)/q1(q2 + q1)
– p1(q1 – q2)/q1(q2 + q1)
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(distribute denominator)
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x1 = (p2 + p1)/(q2 + q1) – p1/q1
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(simplify)
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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).
