Comments on: Circles, circles everywhere

By: A good problem solving problem « Continuous Everywhere but Differentiable Nowhere

Wed, 21 Apr 2010 00:18:03 +0000

[...] down in the comments and how you came up with it (or blog about it), awesome. Just like with this “circles, circles everywhere” problem where someone posted the most elegant solution [...]

By: Carnival of Mathematics #60 « ∑idiot's Blog

Carnival of Mathematics #60 « ∑idiot's Blog — Fri, 04 Dec 2009 05:03:17 +0000

[...] Sam Shah at Continuous Everywhere but Differentiable Nowhere: Circles, circles everywhere [...]

By: jd2718

jd2718 — Tue, 01 Dec 2009 12:32:15 +0000

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…

By: jd2718

jd2718 — Sat, 28 Nov 2009 16:33:08 +0000

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

By: Chris Taylor

Chris Taylor — Thu, 26 Nov 2009 11:18:57 +0000

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).

By: Matt E

Matt E — Wed, 25 Nov 2009 20:10:56 +0000

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.

By: samjshah

samjshah — Wed, 25 Nov 2009 18:35:43 +0000

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!

By: Chris Taylor

Chris Taylor — Wed, 25 Nov 2009 17:59:25 +0000

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.