Selecting two points on two circles such that their distance is greater than one circle's diameter

Given are two, disjoint circles $$k_1, k_2$$ of equal diameter $$d$$ in space. One can then choose two points $$P_1, P_2$$, one on each circle such that the distance between them is greater than $$d$$.

How can this problem be solved "topologically"?

My thoughts thus far:

One can continuously map $$[0,1]$$ onto the circle so that $$0$$ and $$1$$ are mapped to onto same point. Thus, there is a function $$g : [0,1] \times [0,1] \rightarrow k_1 \times k_2$$

Now I am looking at $$S = \{ r \in \mathbb{R} | r = d(a,b) \text{ with } a \in k_1, b \in k_2 \}$$ with $$d(a,b)$$ measuring the euclidean distance between the two points $$a$$ and $$b$$.

I can now look at a map between $$[0,1] \times [0,1] \rightarrow S$$ and I should be able to reshape the figure of my square $$[0,1] \times [0,1]$$, where the edges are glued together, now to a sphere.

As much I feel proud making this connection, as much I feel stuck and stupid. I don't think this gets one anywhere.

How can one salvage this? My 2nd idea revolved around looking at pairs of points on each circle. Each point on a circle can be uniquely identified by its antipodal point. So now I would repeat above's process, however I would define $$S' = \{ r \in \mathbb{R} | r = max(d(a,b), d(a', b')) \text{ with } a, a' \in k_1, b, b' \in k_2 \}$$ and $$a, a'$$ and $$b, b'$$ being antipodal points respectively.

While this all seems nice, I don't see how put everything together. For example, I haven't included the diameter of $$d$$ of the circles.

Does someone see a better, more constructive way to solve this problem?

• This is likely not "topological", but draw a line through the centers of the $2$ circles. Where this line crosses each circle on the opposite side of the other circle's center defines $2$ points which are greater than a distance $d$ from each other. If you put into "topological" terms, it might give you an appropriate answer. – John Omielan Feb 26 at 20:54
• This only works, if both circles lie a the plane^^ – Imago Feb 26 at 20:56
• You are right, it's more complicated than that in $3$ or more dimensions. Good luck with proving the statement in those conditions. – John Omielan Feb 26 at 20:58
• What happens if the circles are on an ellipsoid where distance is measured on the surface? – William Elliot Feb 26 at 22:06

In four or more dimensions, it's false. The circles $$(\cos\theta,\sin\theta,0,0)$$ and $$(0,0,\cos\phi,\sin\phi)$$ in $$\mathbb{R}^4$$ each have diameter $$2$$, and every point in the first circle is at distance exactly $$\sqrt{2}$$ from every point in the second.
In three dimensions, the circles can be represented as $$k_i=\{O_i+v : v\in V_i\text{ and }\|v\|=\frac12d\}$$ where $$O_i$$ is the center and $$V_i$$ is a vector subspace of $$\mathbb{R}^3$$ of dimension $$2$$. Since the sum of the dimensions of $$V_1$$ and $$V_2$$ is greater than $$3$$, they have a nontrivial intersection. Let $$w$$ be a vector of length $$\frac12d$$ in this intersection.
Now, the four points $$A=O_1+w,B=O_1-w,C=O_2-w,D=O_2+w$$ lie on the circles and form a parallelogram. Since the circles are disjoint, they are four different points in $$\mathbb{R}^3$$. We claim that at least one of the diagonal lengths $$\|A-C\|$$ and $$\|B-D\|$$ is strictly greater than $$d=\|A-B\|=\|C-D\|$$.
Why? Apply the "parallelogram law": $$\|A-C\|^2+\|B-D\|^2 = 2\|A-B\|^2+2\|B-C\|^2 > 2\|A-B\|^2=2d^2$$ With the sum strictly greater than $$2d^2$$, one of $$\|A-C\|^2$$ and $$\|B-D\|^2$$ must be strictly greater than $$d^2$$. Take the square root, and we're done.