Construct point on a circle such that the reflection in that point is horiztonal

Let $P$ be a point in the plane outside the unit circle. There is a unique point $Q$ on the circle such that a light ray from $P$ is reflected in the circle at $Q$ and emerges parallel to the $x$-axis. Is it possible to construct $Q$ using a ruler and compass?

If $P$ is a point on the piece of paper in this picture, the image of $P$ will appear to an observer at infinity to be at a point easily constructed from $Q$. I would also be interested in a proof that such a construction is impossible.

• cool picture (+1) – tired Dec 20 '16 at 14:26
• I'm shocked ! WOW!!! – Arman Malekzadeh Dec 20 '16 at 14:58
• – Michael Hoppe Dec 20 '16 at 15:04
• If $P$ corresponds to the complex number $z=x+iy$ with $x>0$ or $y>1$, and $Q$ corresponds to $w$ with $|w|=1$, then you require $$w^2 = \frac{z-w}{|z-w|}.$$ I've no idea if $w$ is constructible. Nice question! – Bob Pego Dec 20 '16 at 18:07
• Take $w=s+it$ in my previous comment, multiply by $|z-w|\bar w^2$ and take imaginary part. Then because $w\bar w=1$ we require $$0=\Im (z\bar w^2-\bar w)=y(s^2-t^2)-2xst+t, \quad s=\sqrt{1-t^2}.$$ Moving $2xst$ over and squaring gives a quartic polynomial in $t$. Typically solving such a polynomial requires taking a cube root, which may not always be constructible. This is not a proof, but it seems probable there is no general construction. – Bob Pego Dec 21 '16 at 17:14

I don't have an answer to your question (which I like a lot!), but I have two thoughts, the first related to @Bob Pego's comment: at a purely algebraic level, the requirement that the normal vector split the ray and its reflection is satisfied by four points on the circle in general (corresponding to Bob's 4th degree equation, and showing that this equation didn't introduce extraneous roots). Here's a (very rough) picture of the situation, where each solution is colored differently: The red dot is P; the four circle-intersections are (I hope) clear.