Circles Surrounding Circles – Part II

Here it is, part II of ‘Circles Surrounding Circles’. Part I can be found here.

Note: While I was researching this topic, I stumbled upon this page. It presented exercises exploring exactly what I wanted to explore. So these posts have basically turned into my commentary on the exercises provided there (Even more so in this post than the last). I have done some original work, but I thought I’d give the credit where it’s due.

Now let us ask the question: “What if the inner circle could be a different size to the outer circle?” For the first case, let us consider the following set-up:

There are 12 outer circles here
There are 12 outer circles here

Let us try and determine the ratio between the inner circle’s diameter and the outer circle’s diameter. As in the last post, we shall look at a triangle whose points are the center’s of two outer circles and the center of the inner circle. This is shown, with labels, below.


Let r be the radius of the inner circle and s the radii of the outer circles, we’re looking for the ratio between these two lengths. For similar arguments shown in the last post, we can deduce that the triangle OPQ is an isosceles triangle with sides PO and QO being equal to (s+r) and side PQ being equal to 2r.

As triangle OPQ is isosceles, we can therefore conclude that triangle POR is congruent to triangle QOR. This means that the angles ORP and ORQ are congruent. They also lie on a straight line, so angles ORP and ORQ add up to 180 degrees. Taking both these facts we can deduce that they are both right angles.

Now we have established that triangles OPR and OQR are both right angled triangles. This is important as it allows us to utilize some trigonometric results. Considering triangle OPR, let us first first the angle POR in order to use the sine function to finally find the ratio between the radii r & s.

In a similar fashion to the idea from part one that only six equalateral triangles can be packed around O. We can deduce angle POR as follows, because there are twelve outer equal circles, the measure of angle POQ is one twelfth of 360 degrees or 30 degrees. Angles POR and QOR are congruent and so they much each be 15 degrees.

Now we have everything we need to use the sine function. The sine of angle POR is the length of the side opposite the angle, |PR| divided by the length of the hypotenuse |OP|. This all means we have the following equation.


Multiplying both sides by r+s and then solving for s in terms of r we find:

Using a calculator we can solve this to find that, s is equal to about 0.349r. This is in line with the picture, which shows that the outer circles are significantly smaller than the inner circle.
We can use a similar approach to evaluate this next case.
I won’t bore you with the in’s and out’s of this case, as it is pretty much identical to the first case. In this case we find that the isosceles triangle looks like so:
Using a similar argument from before we can show that here the angle POR is 45 degrees. This provides the following ratio, where t is the radius of the outer circles and r is again the radius of the inner circle.
Using a calculator once again, we find that t is about 2.41r. This is once again in line with the picture, so we can now be pretty sure we have ourselves a reliable formula describing the relationship between the inner circle and the outer circles. Let us use this formula to shed some more light on our original problem by setting the inner circle radius equal to the radii of the outer circles.
We find that formula is only equal when the angle is 30 degrees (as sin30 is equal to one half, making the term inside the brackets equal to 1). 30 degrees corresponds to the construction of six equilateral triangles we considered in part one. This results seems more satisfying as we have now found a general rule for the problem of circles surrounding circles. Who knew so much was going on when we were sorting coins on a flat surface?
Note: I really cannot take credit for this part II post, as the mathematics, diagrams and formula come from this page. All I’ve done here is explain it in my own words and if I didn’t mention it I would feel like a fraud!

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