A cone through two circles on a sphere

Question

In page 37 of Blaschke's Vorlesungen Über Differentialgeometrie, vol. III, when proving a theorem on Möbius transformations the author says "Denn durch zwei Kugelkreise läßt sich bekanntlich immer ein Kegel zweiter Ordnung legen, ...". In a free translation:

"As is well-known, there is always a second-order cone through two circles on a sphere".

Although I know how to prove the above fact by interpreting the map of the sphere onto itself given by a central projection in terms of a Lorentzian reflection on the space $\mathbb{R}^4$ of homogeneous coordinates, I would be happy to see a more geometric argument which constructs the vertex of such a cone (let's say, some argument that Poncelet, or even Euclid, would find).

Answer 1

Consider the plane passing through the centers of the sphere and of both circles. This plane intersects the circles at diameters $AB$ and $CD$. If lines $AC$ and $BD$ (or $AD$ and $BC$) meet at $V$, then $V$ is the vertex of the requested cone.

That's because triangles $VAB$ and $VCD$ are similar, hence the two circles are what Apollonius called two subcontrary sections of the cone (Book 1, Proposition 5 in "The conics").

You'll have in general two solutions, with $V$ inside or outside the sphere.

EDIT.

Here's the proof (as given by Apollonius) that a subcontrary section to the base of a circular oblique cone is also a circle.

Let's consider then a cone of vertex $V$ and a circular base (see figure below). Diameter $AB$ of the base is chosen so that plane $VAB$ is perpendicular to the plane of the circle.

On triangle $VAB$ choose points $C$ (on $VA$) and $D$ (on $VB$) suche that $\angle VDC=\angle VAB$. The plane through $CD$ perpendicular to triangle $VAB$ cuts then the cone along a curve $CPD$ (which is called a subcontrary section of the cone). We want to show that this curve is a circle.

Indeed, consider any point $P$ on that curve and the plane through $P$ parallel to the base. That plane cuts the cone along a circle $GPI$, with diameter $GI$ lying on triangle $ABV$ and parallel to $AB$. Diameters $CD$ and $GI$ intersect at $H$: by construction $PH$ is perpendicular both to $CD$ and $GI$.

As $PGI$ is a circle, we have: $$ PH^2=GH\cdot IH. $$ But triangles $CGH$ and $DIH$ are similar (because $\angle IDH=\angle VAB=\angle CGH$), hence $$ GH:CH=DH:IH,\quad\text{that is:}\quad CH\cdot DH=GH\cdot IH. $$ Comparing with the previous equality we then get: $$ CH\cdot DH=PH^2. $$ But this holds for any point $P$ on the curve, hence that curve is a circle, QED.