# Circles, circles everywhere

So we’re off for Thanksgiving today (phew!) and after a few really great days, this two-day week was pretty much a bummer. The one fun mathematical thing I’ve worked on is a problem involving geometry. Ew, I know. But I got really into it, and it got me thinking of about a million other extensions, questions, methods of attack, etc. I was thinking in so many different ways — about symmetry and about limiting and degenerate cases and about angles and such.

(1) You have two circles, radius 3 and radius 5, tangent to each other. You want to draw a third circle tangent to the given two circles. In fact, you realize there are an infinite number of these circles. So the first question is: what is the locus of all points which is comprised of the centers of these infinite circles? If you want a small hint, go after the jump for a picture. (In case it wasn’t clear, we are taking about circles which are externally tangent to each other.)

(3) Can you find a third circle tangent to the given two circles such that the centers of the three circles forms a right triangle (if possible).

(4) What if we ask a similar question about spheres? If you are given two spheres of radius a and radius b, what is the locus of all points which is comprised of the centers of spheres tangent to the given two spheres?

So if you are bored over your Thanksgiving holiday, you might want to have some fun with this. I’ve solved the first two. I haven’t had time to think about the third yet, though I know the solution won’t be (too) hard. The fourth one? Eh, I anticipate it to be pretty tough. But having solved the first two will definitely help! [update 5 minutes after posting: Eh, nevermind, I think I know the answer to the fourth one… not really hard!]

1. Chris Taylor says:

Fun little problem! If the circles have radii a and b, the locus is a hyperbola if a and b aren’t equal, or a straight line if they are. In the limit as a or b tend to 0, the locus becomes a parabola. In the case where one of your initial circles is actually inside the other, the locus is an ellipse.

1. samjshah says:

Lovely! Now if you have time could you briefly jot down what approach you took to solving it? Geometric? Algebraic? I’m curious about the way you went about getting the solution. I honestly struggled a lot when I first started doing the problem — but then I went a brute force way and got the solution.

I didn’t think about internally tangent circles, and that’s super nice!

1. Chris Taylor says:

My first approach was algebraic: Let the centres of the three circles be A, B, C with radii a, b and c. I put A at the origin, and B at the point (a+b,0). Letting C be at the point (x,y) you require that the length of the line AC is a+c, and the length of BC is a+b. That gives you two equations

x^2 + y^2 = (a+c)^2

(x-a-b)^2 + y^2 = (b+c)^2

Eliminating c from these equations gives you a quadratic form in x and y: looking at the coefficient of x^2 allows you to determine what kind of conic you have.

After I did this I realised that there was a better solution. Since the lines AC and BC have lengths a+c and b+c, you have the relation

|AC| – |BC| = a – b = constant

which is the geometrical definition of the locus of a hyperbola. In the case where A is inside B, the length of AC is a+c and the length of BC is b-c, so you have

|AC| + |BC| = a + b = constant

which is the geometrical definition of the locus of an ellipse. That you get a straight line when a=b is obvious, and with a little bit of thinking you can work out that you get a parabola when a=0, and a circle when a+b=0 (i.e. when the two circles lie on top of each other).

2. Matt E says:

Did you just “ew” Geometry? For shame. It’s my fave.

You can also extend the problem to starting with any two circles, tangent or not, intersecting or not… and think about the locus of centers of common tangent circles.

As for the hyperbolas & ellipses, if you think about the geometric definition of those shapes (using their foci) it’s relatively simple to see why they are what are generated.

3. jd2718 says:

Fumbling around, I called the new radius r, described a triangle joining the centers with sides:
8, r+3, and r+5, and used the law of Cosines to find the angle at the center of the new circle:

Cos(new angle) = (r^2 + 8r – 15)/(r^2 + 8r + 15)

Didn’t get me an answer, but I thought the symmetry was fairly cool. Horizontal asymptote at 1 shows that THAT angle approaches 0, but what of the vertical asymptotes?

Jonathan

4. jd2718 says:

So, I thought some more, and realized I should have thought less.

The distance from the new center to A is r+3, to B is r+5, and the difference of the distances to A and B is 2.

Locus of all points, the difference of whose distances from 2 given points is a constant…