Transformations of tangent lines into curves

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.

• 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. – 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.
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 $$\phi=\frac{\pi}{2}$$.