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.