Problem of Apollonius

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{{Image Description Ready |ImageName=Apollonian Gasket |Image=Apollonian.jpg |ImageIntro=This an example of a fractal that can be created by repeatedly solving the Problem of Apollonius.

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The problem of Apollonius involves trying to find a circle that is tangent to three objects: points, lines, or circles in a plane. The most famous of these is the case involving three different circles in a plane, as seen in the picture to the left. The given three circles are in red, green, and blue, while the solution circle is in black.

Apollonius of Perga posed and solved this problem in his work called Tangencies. Sadly, Tangencies has been lost, and only a report of his work by Pappus of Alexandria is left. Since then, other mathematicians, such as Isaac Newton and Descartes, have been able to recreate his results and discover new ways of solving this interesting problem.

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The problem usually has eight different solution circles that exist that are tangent to the given three circles in a plane. The given circles must not be tangent to each other, overlapping, or contained within one another for all eight solutions to exist.

Given three points, the problem only has one solution. In the cases of one line and two points; two lines and one point; and one circle and two points, the problem has two solutions. Four solutions exist for the cases of three lines; one circle, one line, and one point; and two circles and one point. There are eight solutions for the cases of two circles and one line; and one circle and two lines, in addition to the three circle problem.

1 A More Mathematical Explanation
- 1.1 Algebraic Method
2 Apollonian Gasket
3 References
4 Mathematica Programs
- 4.1 Ideas for the future

A More Mathematical Explanation

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There are many different ways of solving the problem of Apollonius. The few that are easiest to understand include using an algebraic method or an inverse geometry method.

Algebraic Method

This method only uses math up to the level of understanding quadratic equations. We will proceed by setting up a system of quadratic equations and solving for the radius, r, of the unknown circle.

We start by labeling the center of each of the given circles $(x$ ₁ $,y$ ₁ $)$ , $(x$ ₂ $,y$ ₂ $)$ , and $(x$ ₃ $,y$ ₃ $)$ . We will call the center of the unknown circle $(x,y)$ . $r$ ₁, $r$ ₂, and $r$ ₃ are the different radii of each of the given circles.

From this we are able to write our equations:

$(x - x$ ₁ $)^2 + (y - y$ ₁ $)^2 = (r \pm r$ ₁ $)^2$
$(x - x$ ₂ $)^2 + (y - y$ ₂ $)^2 = (r \pm r$ ₂ $)^2$
$(x - x$ ₃ $)^2 + (y - y$ ₃ $)^2 = (r \pm r$ ₃ $)^2$

Next we are able to expand each of the equations to see better how they can relate to each other.

Expanding gives us:

$(x^2+y^2-r^2)-2xx$ ₁ $-2yy$ ₁ $\pm2rr$ ₁ $+(x$ ₁ $^2+y$ ₁ $^2-r$ ₁ $^2)=0$
$(x^2+y^2-r^2)-2xx$ ₂ $-2yy$ ₂ $\pm2rr$ ₂ $+(x$ ₂ $^2+y$ ₂ $^2-r$ ₂ $^2)=0$
$(x^2+y^2-r^2)-2xx$ ₃ $-2yy$ ₃ $\pm2rr$ ₃ $+(x$ ₃ $^2+y$ ₃ $^2-r$ ₃ $^2)=0$

We can now look at the equations and see how we can subtract them from each other. So we will take the second and third equation minus the first equation.

Second minus first gives us:

$2(x_1-x_2)x+2(y_1-y_2)y+2(\pm r_1 \pm r_2)r=(x_1^2+y_1^2-r_1^2)-(x_2^2+y_2^2-r_2^2)$

Third minus first gives us:

$2(x_1-x_3)x+2(y_1-y_3)y+2(\pm r_1 \pm r_3)r=(x_1^2+y_1^2-r_1^2)-(x_3^2+y_3^2-r_3^2)$

For the sake of simplicity, we'll define some new variables. Let

$a_2=2(x_1-x_2)$ ; $b_2=2(y_1-y_2)$ ; $c_2=2(\pm r_1 \pm r_2)$ ; $d_2=(x_1^2+y_1^2-r_1^2)-(x_2^2+y_2^2-r_2^2)$

$a_3=2(x_1-x_3)$ ; $b_3=2(y_1-y_3)$ ; $c_3=2(\pm r_1 \pm r_3)$ ; $d_2=(x_1^2+y_1^2-r_1^2)-(x_3^2+y_3^2-r_3^2)$

Now our two equations can be written as

$a_2 x+b_2 y +c_2 r=d_2$

$a_3 x+b_3 y +c_3 r=d_3$

Since this is a simple linear system of equations, we can solve it for x and y in terms of r.

Solving the first equation for x:

$a_2x=d_2-c_2r-b_2y \rightarrow x=\frac{d_2-c_2 r -b_2 y}{a_2}$

Substituting that in to the second equation allows us to find y in terms of known values and r:

$a_3\left(\frac{d_2-c_2 r -b_2 y}{a_2} \right)+b_3 y +c_3 r =d_3$

$a_3 d_2-a_3c_2 r -a_3 b_2 y+a_2 b_3 y +a_2 c_3 r =d_3 a_2$

$y(a_2 b_3-a_3 b_2)=a_2 d_3-a_3 d_2 +(a_3 c_2 -a_2 c_3)r$

$y=\frac{a_2 d_3-a_3 d_2 +(a_3 c_2 -a_2 c_3)r}{a_2 b_3-a_3 b_2}$ .

Rather than substituting back in to find x, it is actually simpler to go through the same process we used to find y to get x in terms of known values and r.

First, we solve the first equation for y.

$b_2y=d_2-c_2r-a_2x \rightarrow y=\frac{d_2-c_2 r -a_2 x}{b_2}$

Plugging into the second equation gives us

$a_3 x+b_3\left(\frac{d_2-c_2 r -a_2 x}{b_2}\right) +c_3 r =d_3$

$a_3 b_2 x+b_3 d_2 -b_3 c_2 r-a_2 b_3 x +b_2 c_3 r=b_2 d_3$

$x(a_3 b_2-a_2 b_3)=b_2 d_3-b_3 d_2 +(b_3 c_2-b_2 c_3) r$

$x=\frac{b_2 d_3-b_3 d_2 +(b_3 c_2-b_2 c_3) r}{a_3 b_2-a_2 b_3}$ .

With $x=\frac{b_2 d_3-b_3 d_2 +(b_2 c_3-b_3 c_2) r}{a_3 b_2-a_2 b_3}$ and $y=\frac{a_2 d_3-a_3 d_2 +(a_3 c_2 -a_2 c_3)r}{a_2 b_3-a_3 b_2}$ , we plug in values for the a's, b's and c's (which we get from the original information about the centers and radii of the three circles) to calculate x and y. Using our very first equations for the circles, we can then solve for r.

Let's pick three circles and find the mutually tangent ones. Let's choose some coordinates and radii for our three circles.

Take $(x_1,y_1,r_1)=(0,3,2)$ , $(x_2,y_2,r_2)=(-2,-2,1)$ and $(x_3,y_3,r_3)=(3,-3,3)$ . These three circles are shown below.

Now we want to calculate the a's, b's, and d's first.

$a_2=2(x_1-x_2)=2(0--2)=4$

$a_3=2(x_1-x_3)=2(0-3)=-6$

$b_2=2(y_1-y_2)=2(3--2)=10$

$b_3=2(y_1-y_3)=2(3--3)=12$

$d_2=(x_1^2+y_1^2-r_1^2)-(x_2^2+y_2^2-r_2^2) =(0^2+3^2-2^2)-(2^2+2^2-1^2)=(9-4)-(4+4-1)=-2$

$d_3=(x_1^2+y_1^2-r_1^2)-(x_3^2+y_3^2-r_3^2) =(0^2+3^2-2^2)-(3^2+3^2-3^2)=(9-4)-(9+9-9)=-4$

Calculating the c terms requires a bit more thought since there are the $\pm$ signs. The choice of these signs determines which circle we are solving for. We simply must be consistent in all of our applications of signs for a given r. For the first example, let's simply take all of the plus signs. Then

$c_2=2(r_1+r_2)=2(2+1)=6$

$c_3=2(r_1+r_3)=2(2+3)=10$ .

Now we can calculate x and y for this first circle.

$x=\frac{b_2 d_3-b_3 d_2 +(b_3 c_2-b_2 c_3) r}{a_3 b_2-a_2 b_3}=\frac{(10)(-4)-(12)(-2)+(12(6)-10(10))r}{-6(10)-4(12)}=\frac{-16-(28)r}{-108}$

$x=\frac{1}{27}(4+7r)$ .

$y=\frac{a_2 d_3-a_3 d_2 +(a_3 c_2 -a_2 c_3)r}{a_2 b_3-a_3 b_2}=\frac{(4)(-4)-(-6)(2)+ (-6(6)-4(10))r}{12(4)-(10)(-6)}=\frac{-28-(76)r}{108}$

$y=\frac{-7-19r}{27}$ .

Now we can return to one of our first equations to find r.

$(x-x_1)^2+(y-y_1)^2=(r+r_1)^2$

Note that here the sign of $r_1$ is positive. That is because we took the positive sign when we solved for the c values. Plugging in values, we get

$\left(\frac{1}{27}(4+7r)\right)^2+\left(\frac{-7-19r}{27}-3 \right)^2=(r+2)^2$

This equation is quadratic in r, so it can be solved using the quadratic formula (or a graphing calculator, if you prefer). When the dust clears, we get $r\approx 4.729$ (the other value that comes out of the quadratic formula does not work when plotted).

Now we can plot this circle with center $\left(\frac{1}{27}(4+7(4.729)),\frac{-7-19(4.729)}{27} \right)$ . It is shown below in red, with the original circles in black.

We can see that it is indeed tangent to the three original circles!

This process can be repeated choosing different signs for the different r values in the c coefficients to find the other seven circles.

Apollonian Gasket

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The Apollonian gasket is an example of one of the earliest studied fractals and was first constructed by Gottfried Leibniz. It can be constructed by solving the problem of Apollonius iteratively. It was a precursor to Sierpinski's Triangle, and in a special case, it forms Ford Circles.

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Constructing the gasket begins with three mutually tangent circles. By solving this case of the problem of Apollonius we know that there are two other circles that are tangent to the three given circles. We now have five circles from which to start again.

Repeat the process with two of the original circles and one of the newly generated circles. Again, by solving Apollonius' problem we can find two circles that are tangent to this new set of three circles. Although, we already know one of the two solutions for this set of three circles; it is the other of the three circles that we started with.

Repeating this process over and over again with each set of three mutually tangent circles will create the Apollonian gasket.

References

Math Pages, Apollonius' Tangency Problem

MathWorld, Apollonius' Problem

Wikibooks , Apollonian fractals

Mathematica Programs

Anna created several programs that solve the problem of Apollonius in the case of three non-tangent, non-intersecting circles that the user inputs. The programs were written in Mathematica 7, but are likely compatible with Mathematica 5 and/or 6.

The first program automatically plots all eight solutions. Click here to download this program

The second program allows the user to choose which solutions to plot in groups of two. Click here to download this program

Ideas for the future

AnnaP has a mathematica program that solves the problem for three non intersecting circles. Turning this code into an applet where people could input three circles would be really cool! Email Anna to get the mathematica script if you want to work on this.

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Problem of Apollonius

From Math Images

Contents

A More Mathematical Explanation

Algebraic Method

Apollonian Gasket

References

Mathematica Programs

Ideas for the future

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