What is the probability that the intersection of two random circles on a disk contains the centre of the disk?

Question

On a disk, two independent uniformly random points are chosen.

Two circles, $C_1$ and $C_2$, pass through these points and are tangent to the perimeter of the disk.

What is the probability that the intersection of $C_1$ and $C_2$ contains the centre of the disk?

I will post my answer, showing that the probability is $1/2$.

My answer uses integration; I am looking for a more intuitive explanation (here is an example of an intuitive explanation). If no such explanation is found, then I'll add this to my list.

I made a random circle generator.

Context

This question was inspired by a little geometry puzzle.

Answer 1

The probability is $1/2$

Assume that the radius of the disk is $1$.
Let the first random point be $P$, at a distance $x$ from the centre of the disk.

Claim: The intersection of $C_1$ and $C_2$ contains the centre just if the second random point lies in the shaded region.

Proof of claim: If the second random point is on one of the six blue arcs, then one of the $C$ circles passes through the centre of the disk. From each blue arc, the second random point can move $(1)$ into the blue circle, or $(2)$ out of the blue circle. For each of the twelve movements (or six, applying symmetry), we check whether this results in both $C$ circles containing the centre. The claim follows.

The unshaded region is comprised of a circular sector (bounded by the dashed lines) and two semidisks of radius $\frac12$, so its area is $\arccos x+\frac{\pi}{4}$. We integrate this area from $x=0$ to $x=1$, noting that the probability density function of $x$ is $f(x)=2x$.

$P(\text{not contain})=\dfrac{1}{\pi}\displaystyle\int_0^12x\left(\arccos x+\dfrac{\pi}{4}\right)dx=\dfrac12\space$ (integration by parts)

$\therefore P(\text{contain})=1-\dfrac12=\boxed{\dfrac12}$

Simulations

I did a simulation: $10^7$ trials yielded a proportion of $0.50002$ containing the centre.

Incidentally, a simulation suggests that the expected area of the five regions formed by the two circles are all equal, i.e. $1/5$ the area of the disk.

More importantly

The diagram in this answer reminds me of a cartoon character that I cannot recall exactly. The shaded area is perhaps the character's hair, or a hat. Can anyone identify the character?

Answer 2

An attempt to explain probability might perhaps differ from 1/2.

see https://www.geogebra.org/classic/x8mzfc2y

Let Bc be center of outer blue circle. Let Bg be center of inner green circle. Let Br be center of inner red circle.

Let cyan BG be where blue and green circles touch. Let magenta BR be where blue and red circles touch.

Since green and red circles intersect, at least one of them, say green one, have radius more than half the radius of the blue circle.

If Mg is middle of Bc:BG Gc is on Mg:Bc.

Rc is on BR:Bc.

In geogebra I take BG fixed and let BR move on blue circle and let Gc move and Rc move.

I see that Bc enters intersection when Rc passes Mr which is middle of Bc:BR.

That hints to 1/2 probability.

But if Rc:BR is small, there is no green red intersection and there are no A B so part of the segment Bc:BR seems to not apply...

Of course, we talk about uniform random points A B common to green and red circle.

Perhaps my choices do not translate to such uniform distribution.

I apologize for your valuable time.