# 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.

• If $P$ and $Q$ lie on a diameter of $D$, then there is no such circle. – TonyK Aug 2 at 12:21
• @TonyK: In that case we can think the line $PQ$ as a circle with infinite radious. – Fabio Lucchini Aug 2 at 12:23
• I think, a way to prove uniqueness is using Ptolemy's theorem: en.wikipedia.org/wiki/Ptolemy%27s_theorem – Βασίλης Μάρκος Aug 2 at 13:38

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.

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.