Length of a Tangent

circles1I’ve seen variations of this one a few times, so I thought I’d give it a quick write-up. The simpler version is: Given two circles that are tangent and a line that is cotangent to them, what is the length of the segment between the points of tangency?

To make things easier, the radii are usually given, as 9 and 4 for instance. However, I’m going to keep them as unknowns in order to derive a general formula.

First of all, let’s add a few segments. Point E is the point of tangency between the circles; \(\overline{AC}\) will always go through that point, and \(\overline{AE}\) and \(\overline{CE}\) are radii of the circles. Meanwhile, point F is located so that BDCF is a rectangle.

circles2

Now we’ve created a right triangle. One leg of that triangle, CF, is a base of the rectangle, and so is the same length as BD. We want to solve for BD in terms of the two radii, so let us instead solve for CF, one leg of the right triangle.

AC, the hypotenuse of the right triangle, is equal to the sum of the radii, that is, \(a + b\). Note that BF and CD are also the same length. Hence AF, the other leg of the right triangle, is equal to the difference of the radii, \(a – b\). By the Pythagorean Theorem, \(AF^2 + CF^2 = AC^2\) and so \(CF = \sqrt{AC^2 – AF^2}\).

The radicand is \((a + b)^2 – (a – b)^2 = a^2 + 2ab + b^2 – a^2 + 2ab – b^2 = 4ab\), so \(CF = \sqrt{4ab} = 2\sqrt{ab}\). Thus, as a general rule, the length of a segment between points of tangency on tangent circles is equal to twice the square root of the product of their radii (i.e., \(c = 2\sqrt{ab}\)).

circles3We can use this to develop a formula for a more complicated situation. Let’s say we have three circles that are mutually tangent and which are cotangent to a line, as shown in the diagram. If we know the radii of the two larger circles, what is the radius of the third circle?

We use the formula three times:

  • \(BD = 2\sqrt{ab}\)
  • \(BH = 2\sqrt{ah}\)
  • \(HD = 2\sqrt{bh}\)

Because \(BH + HD = BD\), \(2\sqrt{ah} + 2\sqrt{bh} = 2\sqrt{ab}\) and therefore \(\sqrt{ah} + \sqrt{bh} = \sqrt{ab}\). If we know two of the values, we can use that to find the third.

We can also solve this for h algebraically:

\[\sqrt{ah} + \sqrt{bh} = (\sqrt{a} + \sqrt{b})\sqrt{h} = \sqrt{ab} \\ \Rightarrow \sqrt{h} = \frac{\sqrt{ab}}{\sqrt{a} + \sqrt{b}} \\ \Rightarrow h = \frac{ab}{a + 2\sqrt{ab} + b} \]

If you prefer to avoid having a radical in the denominator, multiply by the complement \(\left(a + b – 2\sqrt{ab}\right)\) and simplify: \[h = \frac{ab}{a + 2\sqrt{ab} + b} = \frac{ab(a + b – 2\sqrt{ab})}{(a + b)^2 – 4ab} \\ = \frac{a^2b + ab^2 – 2ab\sqrt{ab}}{a^2 + b^2 – 2ab}\]

Compare this to what we get using the special case of Descartes’ Theorem: \[\frac{1}{h} = \frac{1}{a} + \frac{1}{b} \pm \frac{2}{\sqrt{ab}} \\ = \frac{b\sqrt{ab}}{ab\sqrt{ab}} + \frac{a\sqrt{ab}}{ab\sqrt{ab}} \pm \frac{2ab}{ab\sqrt{ab}} \\ =\frac{\pm 2ab + (a + b)\sqrt{ab}}{ab\sqrt{ab}} \\ \Rightarrow h = \frac{ab\sqrt{ab}}{\pm 2ab + (a + b)\sqrt{ab}}  \]

Above, we derived \(h = \frac{ab}{a + 2\sqrt{ab} + b}\). Multiply the top and bottom by \(\sqrt{ab}\) to get \[h = \frac{ab\sqrt{ab}}{\sqrt{ab}(a + 2\sqrt{ab} + b)} \\ = \frac{ab\sqrt{ab}}{2ab +(a + b)\sqrt{ab}} \]

This is one of the solutions found using Descartes’ Theorem. The other solution to Descartes’ Theorem yields a larger circle. For instance, given radii of 4 and 9, Descartes’ theorem yields values of 36/25 and 36; we only want the smaller value.

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