If two $n$-spheres $S^n$ are made to be tangent, then what is their intersection geometrically? Will it simply be a point or some other geometric object (also, is it independent of the embedding space)? I ask because I am not too sure how these objects behave with respect to each other when embedded in $\mathbb{R}^{n+1}$ for $n\ge 3$.
Thanks in advance!
On
If the spheres are $(n-1)$-spheres in $n$ space and are externally tangent, transforming linearly, we can find a coordinate system where one is centred on the origin with radius $r_{a}$, the other is centred at $(1, 0, \ldots, 0)$ with radius $r_{b}$ and with $r_{a} + r_{b} = 1$.
Here, the equations of the spheres are $$ \sum_{k=1}^{n} x_{k}^{2} = r_{a}^{2} $$ and $$ (x_{1} - 1)^{2} + \sum_{2}^{N} x_{k^{2}} = r_{b}^{2} $$
We can write the tangent relationship as $$ \sqrt{\sum_{k=1}^{N} x_k^2} + \sqrt{(x_{1} - 1)^{2} + \sum_{2}^{N} x_{k^{2}}} = r_{a} + r_{b} = 1 $$
Working this through leads to a single solution of $x_{1} = r_{a}$ for the point of intersection and $x_{k} = 0$ for $k > 1$.
I guess the argument would be similar for internally tangent spheres.
Two distinct $n$-spheres in $\Bbb R^{n+1}$ which are tangent (meet at a point and have a common tangent hyperplane at that point) meet only at that point $P$. To see this, note that their centres $O$ and $O'$ have the property that $O$, $O'$ and $P$ are collinear. So up to Euclidean congruence the centres are $(a,0,\ldots,0)$, $(b,0,\ldots,0)$ and $P$ is $(0,\ldots,0)$ and the radii are $|a|$ and $|b|$. Using the triangle equality, or considering the equations of the spheres shows there are no points in common save $P$ on the spheres.