Construction of a circle through two interior points of a disk and meeting the boundary at two antipodal points 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.
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 $\sigma:\&gt;\dot{S^2}\to{\mathbb R}^2$ 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.
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 $\epsilon$  { Sheaf 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 Sheaf of lines induced by the { Sheaf 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.