0
$\begingroup$

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:

$x^2+y^2=1$

and the line described by:

$x=-1$

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

$x^2+y^2=1$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $\phi=\frac{\pi}{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.