Construction of a circle through two interior points of a disk and meeting the boundary at two antipodal points

Question

Let $D$ be a disk, $P,Q$ be two distinct points in its iterior. A geogebra experiment shows that there exists one and only one (possible degenerated, with infinite radious) circle through $P$ and $Q$ and meeting the boundary of $D$ at two antipodal points.

How to prove it?

Motivation. I'm looking for a model of the real projective plane in the unit disk $D$ (with antipodal points identified). By taking as lines of this models the arcs of circles which meets the boundary of $D$ at two antipodal points, we are proving that given two points $P,D$ one and only line of this model passes through them.

A rule-compass construction. Let $S,T$ be the intersection of $PQ$ with the boundary of $D$, $S',T'$ be the respective antipodal points.

Given two points $X,Y$, let $b_{XY}$ denote the line bisector of the segment $XY$.

Consider $G=b_{T'P}\cap b_{QT}$, $H=b_{SP}\cap b_{QS'}$. Then $I=GH\cap b_{PQ}$ is the center of the required circle.

Answer 1

Let $D$ be the unit disc in ${\mathbb R}^2$. Embed ${\mathbb R}^2$ via $(x,y)\mapsto(x,y,0)$ into ${\mathbb R}^3$, and consider the stereographic projection $\sigma:\>\dot{S^2}\to{\mathbb R}^2$ from the north pole $(0,0,1)$. This map keeps the points of $\partial D$ fixed, and maps the set of circles in $S^2$ to the circles and lines in ${\mathbb R}^2$. A circle $\ne\partial D$ in $S^2$ is a great circle iff it intersects $\partial D$ in two antipodal points.

Now let $\hat P=\sigma^{-1}(P)$ and $\hat Q=\sigma^{-1}(Q)$. There is exactly one great circle $\gamma$ through $\hat P$ and $\hat Q\in S^2$. It follows that $\sigma(\gamma)$ is the single circle through $P$ and $Q$ that intersects $\partial D$ in two antipodal points.

Answer 2

Since you already have a method of constructing the needed circle $C$, I assume your question is about uniqueness. Here is an informal analytic approach that could be formalized.

Let $M$ be the center of $D$, and $N$ be the center of $C$. Since $C$ passes through $P$ and $Q$, $N$ must lie on the line $\ell = b_{PQ}$. Let $X$ be the center of the center of the circle through $P$ and $Q$ which is tangent to $D$ on opposite side of $\overline{PQ}$ from $M$, and let $Y$ be the center of the circle which is tangent to $D$ on the same side of $\overline{PQ}$ from $M$. Consider the direction from $X$ to $Y$ to be up.

Let $Z$ be a point on $\ell$ and consider the circle centered at $Z$ passing through $P$ and $Q$, and the angle formed by the intersections of that circle with $D$ and $M$. Let $\theta$ be the measure on the upper side of that angle. The intersection points will be antipodal if and only if $\theta = \pi$. Let $\theta_0$ be the measure of the upper side of $\angle SMT$ (where, as in the OP, $S$ and $M$ are the intersections of the line through $P$ and $Q$ with $D$.) Because the midpoint of $\overline {PQ}$ is below $M$, $\theta_0 > \pi$.

When $Z = X, \theta = 2\pi$. As $Z$ descends, the intersection points move up the sides of $D$, with $\theta$ decreasing. As $Z$ goes to $-\infty$, the intersection points approach $S$ and $T$, and $\theta \to \theta_0$ from above. Since $\theta > \theta_0 >\pi$, No circle in this region has antipodal intersection points.

When $Z$ is between $X$ and $Y$, its circle does not intersect $D$. When $Z = Y, \theta = 0$, and as $Z$ rises, the intersection points descend the sides of $D$ and $\theta$ increases continuously. As $Z \to \infty, \theta \to \theta_0-$, so at some point $\theta$ must rise past $\pi$. By the intermediate value theorem, there is a point where it equals $\pi$. But since $\theta$ is strictly increasing in $Z$, it cannot be $\pi$ at more than a single point.

Answer 3

Based on the work done in the 1st answer to this question (by Christian Blatter) , we can go further and construct the arc of circle through P and Q by defining the model of the real projective plane as the stereographic projection of the sphere with antipodal points identified. Please find attached below some more details and diagrams.

Let us define the circle model of $\mathbb{R}P^2$ as the the circle D2 obtained via stereographic projection of $S^{2}$ modulo the equivalence relation that identifies antipodal points. Stereographic projection sends lines into lines and circles into circles so that the geodesics on D2 are arcs of circle which go through antipodal points with respect to the center of D2 (Figure 1)
1
Given two points D and E not belonging to $\partial D2$ we want to find the geodesic { C,D,E,B } going through them (Figure 3 below) : 2

So let D and E be points belonging to circle Y (Figure 4) and pick another circle (W) which $\epsilon$ { pencil of circles through D and E } and which intersects Y in the points G and F.

Draw the line DE and the line GF , let I = GF $\cap$ DE, then I is the center of a pencil of lines induced by the { pencil of circles through D and E }. Draw also the line AI which intersects Y in 2 antipodal points $A_{1}$ and $A_{2}$ (not showed), then the circumference Z ={ $A_{1}$DE $A_{2}$ } is the searched unique geodesic.