For which values of $a$ does the hyperboloid $x^2+y^2-z^2=1$ intersect the sphere $x^2+y^2+z^2=a$ transversally? What does the intersection look like for different values of $a$?
What I can see is that if $\sqrt a<1, $ then the intersection is empty and hence transverse. If $\sqrt a=1\iff a=1$, then the intersection points lie on the circle $x^2+y^2=1$ on the $z=0$ plane. I guess at those points the tangent spaces to the hyperboloid and the sphere coincide (they are 2-dim planes). Is that correct? How do I show it more rigorously? (I can use local parametrizations of those manifolds and then compute the image of the corresponding differentials, but this seems to be a huge hassle.) As for the case $\sqrt a > 1$, I guess here the intersection is transverse, but again, how do I show it?
For $a=1$ we obtain $$x^2+y^2=1\\z=0$$therefore the intersection is the unit circle on XY plane centered at (0,0). Generally for $a>1$ we have $$z^2+1=a-z^2$$ therefore$$z=\pm \sqrt {a-1\over 2}\\ x^2+y^2={a+1\over 2}$$
Also for a 2 sheet hyperboloid for example for $x^2+y^2-z^2=-1$ we must have $$z^2-1=a-z^2$$ or $z^2={a+1\over 2}$ in which $a\ge -1$. If so, we have $$x^2+y^2=z^2-1={a-1\over 2}$$which is possible only if $a\ge 1$.
Conclusion: the intersection of the two planes contains two unit circles parallel to XY plane centered at $(0,0,\pm\sqrt{a-1\over 2})$ in the former case and $(0,0,\pm\sqrt{a+1\over 2})$ in the latter case
Here is an image (simulated by MATLAB) indicating the two sheet hyperboloid case: