# A problem with four circles and a square

I was doing random things when I noticed something which seemed strange to me.

What I did was the following.

• Take two points $$A$$ and $$C$$ of the plane. We denote $$\ell$$ the distance $$AC$$.

• Draw a circle $$\mathscr C_0$$ of center $$A$$ and radius $$r_0$$.

• Then define $$M_0$$ to be a point of $$\mathscr C_0$$, and $$M_1$$ the middle of $$[M_0C]$$.

• Finally, define $$M_2$$ and $$M_3$$ such that $$M_0M_1M_2M_3$$ is square.

It should look like something like this:

Then we are interested in the trajectory of $$M_1$$, $$M_2$$ and $$M_3$$ when $$M_0$$ move along the circle $$\mathscr C_0$$.

We notice that every $$M_i$$ seems to move on a circle $$\mathscr C_i$$ of a unique radius $$r_i$$.

But I don't get why this would be true, which leads us to the first questions:

Question 1. Are all trajectories $$\mathscr C_i$$ circles?

Question 2. What are the radius $$r_i$$ in terms of $$\ell$$ and $$r_0$$?

Question 3. Where are located the centers of those circles?

What I noticed is that if the radius $$r_0$$ changes, we still get three other circles, and they are all concentric:

And this is the case even when $$r_0\geqslant \ell$$:

It looks like this if $$r_0$$ varies continuously:

What I did to try to find the centers (since all three circles seems to have the same three centers for different radius $$r_0$$) is to see what it would look like with $$r_0$$ really small:

So the centers seems to be:

• the middle point $$A_1$$ of $$[AC]$$,

• the two points $$A_2$$, $$A_3$$ such that $$AA_1A_2A_3$$ is a square.

Hint

Let $A$ be the origin in the complex plane, and let $C$ be the point $c\in \mathbb{R}$

Let $M_0=re^{i \theta}$ so that $$M_1=\frac 12(c+M_0)$$

Then $$M_2=M_1+i(c-M_1)$$ and $$M_3=M_0+i(M_1-M_0)$$

Now you can obtain the parametric equations of the loci of $M_{1,2,3}$

Consider that vector ${CM_0} = {CA}+{AM_0}$, a combination of a fixed vector and a rotating constant-length vector.
Then:
${CM_1} = \frac 12{CA}+\frac 12{AM_0}$
${CM_2} = {CM_1} +{CM_1}^\perp = \frac 12({CA}+ {CA}^\perp) +\frac 12({AM_0} +{AM_0}^\perp)$
${CM_3} = {CM_0} +{CM_1}^\perp = ({CA} + \frac 12 {CA}^\perp) + ({AM_0} +\frac 12{AM_0}^\perp)$

So at the other points of the square we have in each case a fixed vector to the centre of the circle and a rotating constant-length vector to a point on the circle. Due to the combination of perpendicular vectors we should see $M_2$ and $M_3$ with shifted phase by $45°$ and about $26°$ respectively, which is borne out by your graphics.