I was playing with two rings today, and put one inside the other. It made me think about how many points of a circle touch another circle that is smaller, given that the farthest point of the smaller circle in a given direction touches the farthest point of the bigger circle in the same direction.

Given a tangent line to a circle, the line is, well, tangent. Let's say you have the circle described by:


and the line described by:


Obviously, those are tangent. Now, given the same circle:


but instead of a line, another circle:

$(x-1)^2+y^2 = 2^2$

Isn't it feasible to say that larger circle should have more than one intersection with the smaller circle, because given:

$L=$ distance between line and horizontally-aligned portion of smaller circle.

$C=$ distance between left half of bigger circle and horizontally-aligned portion of smaller circle.

$C$ is always less than $L$ except when $L=0$. Why can't a value of $C$ that is related to a value $L$ (related in the sense that they represent the distance from their assigned object to the smaller circle on the some vertical level) when $L$ is incredibly small be equal to $0$?

If I were to guess without doing any math, I would say $C=Lm$ where $m$ is some function or multiple that is less than zero, and changes the further the vertical distance from the x-axis.

  • $\begingroup$ Wouldn’t your basic argument also apply to the larger circle and its tangent line? That aside, in short the reason is that real numbers are dense: between any two real numbers there’s always another real number, so the distances can’t “collapse” to zero no matter how small they get. $\endgroup$ – amd Sep 29 '18 at 21:18

We can model these points by parametric equations. First of all, we can shift everything a unit to the right to make things a bit easier. Here's how: $x=-1$ becomes $x=0$, $x^2+y^2=1$ becomes $(x-1)^2+y^2=1$, and $(x-1)^2+y^2=4$ becomes $(x-2)^2+y^2=4$.

Then, let $\phi\in[0,\frac{\pi}{2}]$. Thus, $\alpha_{1}(\phi)=(2(1-\sin \phi),2\cos\phi)$ gives a point on the upper-left quarter of the larger circle. Then $\alpha_{2}(\phi)=(1-\sin\phi,\cos\phi)$ gives a point on the smaller circle corresponding to the same angle $\phi$.

If I understand your problem correctly, your problem becomes finding out the distance between each point and the line $x=0$, and seeing if the points overlap more than once.

Short answer: they don't.

Long(er) answer: The points overlap only once because for all $\phi\in[0,\frac{\pi}{2})$, $||\alpha_{1}(\phi)||>||\alpha_{2}(\phi)||$. But then, at $\phi=\frac{\pi}{2}$, $||\alpha_{1}(\phi)||=||\alpha_{2}(\phi)||=0$, where $||u||$ denotes the distance between point $u$ and $(0,0)$. It is true that the points get infinitely close together as $\phi$ approaches $\frac{\pi}{2}$, but due to the density of the Euclidean plane, they only overlap each-other when $\phi=\frac{\pi}{2}$.


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