Two different circles in the plane with nonempty intersection

[Arnold1]

Hi.

Here is a problem I've been trying to solve for some time now. Maybe you could help me.
We have two sets
$$\mathcal {Q}$$ is a set of those circles in the plane such that for any $$x \in \mathbb{R}$$ there exists a circle $$O \in \mathcal {Q}$$ which intersects $$x$$ axis in $$(x,0)$$.$$\mathcal {T}$$ is a set of those circles in the plane such that for any $$x \in \mathbb{R}$$ there exists a circle $$O \in \mathcal {T}$$ which is tangent to $$x$$ axis in $$(x,0)$$.

We need to show that in each of these sets there exist at least two different circles whose intersection isn't empty.
It seems obvious that $$card (Q) \ge card (\mathbb{R})$$. Maybe we could somehow identify each circle with a different rational number?

 

[Evgeny.Makarov]

Maybe we could somehow identify each circle with a different rational number?

Yes, every circle (including interior) contains a point with rational coordinates, just like every segment of the x-axis contains a point with a rational x-coordinate. Therefore, there is at most countably many disjoint circles on a plane.

 

[Arnold1]

So this is it? There are only countably many disjoint circles meeting the above specified conditions nut uncountably many points on x axis. Can we already deduce that at least two circles intersect?

 

[Evgeny.Makarov]

So this is it? There are only countably many disjoint circles meeting the above specified conditions nut uncountably many points on x axis. Can we already deduce that at least two circles intersect?

Yes, we can. If the circles don't intersect, then there is at most countably many of them. But each circle has at most two intersection points with the x-axis, so the number of intersection points is also countable, a contradiction.

 

[blue_raver22]

Hello! It's great that you've been working on this problem and seeking help. I can offer some insights and suggestions.

Firstly, I would like to clarify that \mathcal {Q} and \mathcal {T} are not just sets of circles, but sets of circles with specific properties. This is important to consider when trying to solve the problem.

To show that there exist at least two different circles in each set with nonempty intersection, we can use the fact that the x-axis is a straight line and any circle that intersects or is tangent to it must have a specific equation. For example, a circle in \mathcal {Q} can be represented by the equation (x-a)^2 + y^2 = r^2, where a is the x-coordinate of the center and r is the radius. Similarly, a circle in \mathcal {T} can be represented by the equation (x-a)^2 + (y-r)^2 = r^2.

Using this knowledge, we can see that for any x \in \mathbb{R}, there are infinite circles in \mathcal {Q} and \mathcal {T} that intersect or are tangent to the x-axis. Therefore, it is true that card(\mathcal {Q}) \ge card(\mathbb{R}) and card(\mathcal {T}) \ge card(\mathbb{R}).

However, simply identifying each circle with a different rational number may not be enough to prove the existence of two different circles with nonempty intersection. We would need to show that there exists a rational number for every circle in each set, and that the rational numbers associated with the circles with nonempty intersection are different.

I suggest trying to use geometric arguments and equations to prove the existence of two different circles with nonempty intersection in each set. You may also want to consider the relationship between the sets \mathcal {Q} and \mathcal {T} and how they can help prove the desired result.

I hope this helps and good luck with your problem!