Converting comments to answer, as requested:
The Dandelin spheres answer question (1): a focus of a conic section is the point of tangency of its plane with one of those spheres. Clearly, the point on tangency lies on the cone axis if and only if the plane is perpendicular to that axis; therefore, the axis contains a focus in, and only in, the case of a circle. The axis point has some relation to the conic section, but not one as interesting or useful as a focus. (As for "Does the diameter of the base of the cone pass through the focus of the hyperbola?": note that the cone has no base. The thing extends, and expands, infinitely-far.)
This notion of points seeming to move "out to infinity and come back on the other side" is not uncommon in analytical geometry. (It's kinda what happens along asymptotes, when you think about it.) You might be interested in studying Projective Geometry, which adds a "line at infinity" to the standard Euclidean plane; this broader context helps unify all the conic sections into a single kind of curve that has various different relations to that line. (The parabola, in particular, has its "second vertex" on that line.)
If the given conic is either an ellipse or a hyperbola, vertex $V$ of the cone must lie on the plane perpendicular to the plane of the conic and passing through its foci.
Let $A$ and $B$ be the endpoints of the transverse axis of the conic, $V$ the cone vertex, and $2u$ the aperture angle of the cone. Let then $P$ be any point on the ellipse and $H$ its projection onto $AB$. A plane through $P$, perpendicular to the axis of the cone, intersects the cone along a circle $A'B'P$ (see diagram below), where $A'$ and $B'$ lie on plane $VAB$.
By the intersecting chords theorem we know that $PH^2=A'H\cdot B'H$. But, on the other hand, we have by similitude:
$$
A'H:AH=BC:AB,
\quad\hbox{that is:}\quad
A'H={BC\over AB}\cdot AH;
$$
$$
B'H:BH=AD:AB,
\quad\hbox{that is:}\quad
B'H={AD\over AB}\cdot BH.
$$
If we set $n=VA$, $m=VB$ and $2a=AB$, the above formulas can be written as
$$
A'H={m\sin u\over a}\cdot AH,
\quad
B'H={n\sin u\over a}\cdot BH,
$$
and inserting these into the formula for $PH^2$ we get:
$$
PH^2={mn\sin^2 u\over a^2}\,AH\cdot BH.
$$
If $H$ is a focus of the conic, then $PH={b^2/a}$ is the semi-latus rectum,
while $AH\cdot BH=|a^2-c^2|=b^2$ (as usual $2b$ is the length of the conjugate axis and $2c$ is the distance between foci) and from the above equation we get:
$$
\tag{1}
mn={b^2\over \sin^2 u}.
$$
Another equation for $m$ and $n$ can be found from the cosine rule applied to triangle $AVB$:
$$
\tag{2}
m^2+n^2\mp2mn\cos2u=4a^2,
$$
where sign $-$ must be taken for an ellipse ($\angle AVB=2u$) and sign $+$ for a hyperbola ($\angle AVB=\pi-2u$).
If the conic is an ellipse one obtains from $(1)$ and $(2)$
$$(m-n)^2=4(a^2-b^2),$$
that is the difference of the distances of $V$ from $A$ and $B$ is constant.
The locus of $V$ is then a hyperbola, having its foci at the vertices of the ellipse and semi-major axis $\sqrt{a^2-b^2}$. This hyperbola thus passes through the foci of the ellipse.
If the conic is a hyperbola you get instead
$$(m+n)^2=4(a^2+b^2)$$
and the locus of $V$ is an ellipse having its foci at the vertices of the hyperbola and semi-major axis $\sqrt{a^2+b^2}$. This ellipse thus passes through the foci of the hyperbola.
Finally, if the conic is a parabola of vertex $A$ and semi-latus rectum $p$
one gets from a similar reasoning:
$$
AV=n={p\over2\sin^2u}.
$$
From that it follows that the distance of vertex $V$ from the plane of the parabola is
$$VK=n\sin2u=p\cot u,$$
while the distance from the projection of $V$ on that plane to the focus of the parabola is
$$KF=n\cos2u+p/2={p\over2}\cot^2u.$$
Hence $KF=VK^2/(2p)$ and the vertex thus lies on a a parabola, equal to the given parabola and having its vertex at the focus of the latter.
We can summarise the above results as follows:
the locus of $V$ is a conic section, having center and transverse axis
in common with the given conic but lying in an orthogonal plane, with
eccentricity reciprocal to that of the given conic and vertices at the
foci of the given conic.
You can see below an example: the green ellipse on plane x-y has vertices $A$, $B$, foci $F$, $G$ and eccentricity $1/\sqrt2$. The locus of the vertex $V_1$ of any cone intercepting the ellipse is the pink hyperbola on plane x-z, with vertices $F$, $G$, foci $A$, $B$ and eccentricity $\sqrt2$. Conversely, the ellipse is the locus of the vertex $V_2$ of any cone intercepting the hyperbola. Note also that the axes of the cones are tangent to the locus where the vertex lies.
(from wikipedia)
Best Answer
Viewing a vertical hyperbola in a direction perpendicular to the cutting plane, the "edges" of the cone appear as the hyperbola's asymptotes.
Consequently, the diameter of the circle in question is simply how "wide" those asymptotes are at the level of the hyperbola's vertex. That is, if $A$ is a vertex of the hyperbola with center $O$, and $A^\prime$ is the point where the perpendicular to $\overline{OA}$ at $A$ meets an asymptote, then $|AA^\prime|$ is the radius of the circle. This distance is precisely the length of the conjugate semi-axis.
Writing $a$, $b$, $c$ for the transverse semi-axis, the conjugate semi-axis, and the distance from center to focus, and writing $e$ for the hyperbola's eccentricity, we know that $$a^2 + b^2 = c^2 \quad\to\quad b^2 = c^2 - a^2 = a^2\left(\frac{c^2}{a^2}-1\right) = a^2 \left(e^2-1\right)$$
So, the horizontal circle that meets a vertical hyperbola at its vertex has radius $b = a \sqrt{e^2-1}$. $\square$