Where does a great circle intersect the hypersurface $X_1^2 + X_2^2 = \delta^2 < 1$ of the unit sphere? I have reduced a small problem that I am working on (to do with 2-dimensional minimal cones in arbitrary codimension) to some elementary spherical geometry, which I cannot easily manage or visualise.
Let $S^n$ denote the usual unit sphere in $\mathbb{R}^{n+1}$.
For some fixed $\delta \in (0,1)$, we consider the hypersurface $M := \{X \in S^n : X_1^2 + X_2^2 = \delta^2\} $.
Fix a point $X_0 \in M$ and consider a great circle $\gamma$ passing through $X_0$ whose initial velocity vector is tangent to $M$ at $X_0$.
What does the set $\gamma \cap M$ look like? 
I don't think that an open portion of $\gamma$ can be wholly contained in $M$. But can $\gamma \cap M$ even consist of more than two points? My conjecture is `no' and that $M \cap \gamma = \{X_0,-X_0\}$. 
 A: When $n = 2$, $M$ is a union of two non-great circles and $\gamma$ is never contained in $M$ (but, as you say, intersects $M$ at two antipodal points).
Generally (i.e., for $n \geq 2$), $M$ is the product of $S^1$ (of radius $\delta$) and $S^{n-2}$ (of radius $\sqrt{1 - \delta^2}$), with radii measured in the Euclidean sense (i.e., extrinsically, in $\mathbf{R}^{n+1}$).
Offhand, it looks like your guess (two antipodal points) is right. (Both $M$ and $\gamma$ are symmetric under the antipodal map, so the intersection is as well.) I'm pressed for time, but will try to write up a more detailed answer later (unless someone beats me to it :)
Edit: My offhand assessment above is wrong when $n \geq 3$. (I also didn't think it would take so long to find free time, and hope my initial post didn't forestall others from answering.)
Orthogonally decompose $\mathbf{R}^{n+1}$ as $\mathbf{R}^2 \times \mathbf{R}^{n-1}$, and let $\mathbf{u}=(\mathbf{u}_1, \mathbf{u}_2)$ (etc.) denote the corresponding components of a vector $\mathbf{u}$ in $\mathbf{R}^{n+1}$. If $\mathbf{u}_1$ and $\mathbf{v}_1$ are orthogonal vectors of length $\delta$ in $\mathbf{R}^2$, and if $\mathbf{u}_2$ and $\mathbf{v}_2$ are orthogonal vectors of length $\sqrt{1-\delta^2}$ in $\mathbf{R}^{n-1}$, then
$$\gamma(t) = (\cos t)\mathbf{u} + (\sin t)\mathbf{v}$$
is a great circle (since $\mathbf{u}$ and $\mathbf{v}$ are orthogonal unit vectors) lying in $M$ (decompose $\gamma(t)$ in $\mathbf{R}^2 \times \mathbf{R}^{n-1}$).
Conversely, if a curve of the above form lies in $M$, decompose the vectors $\mathbf{u}$ and $\mathbf{v}$. Since
$$\gamma_i(t) = (\cos t)\mathbf{u}_i + (\sin t)\mathbf{v}_i$$
has constant norm for $i=1, 2$, the fact that $\gamma_i(t) \cdot \gamma_i'(t) = 0$ shows $\mathbf{u}_i$ and $\mathbf{v}_i$ are orthogonal vectors of the same length.
(Note that this argument doesn't contradict the situation when $n = 2$, because the $0$-sphere doesn't contain two orthogonal vectors of the same length.)
