One can use a list of equations to determine the property you require. Note, however, that in many cases it is easier to use derived value(s) as opposed to using equations that rely solely on the coefficients of the equation in general quadratic form. Below are examples of equations one can use.
Properties of an ellipse from equation for conic sections in general quadratic form
Given the equation for conic sections in general quadratic form:
$ a x^2 + b x y + c y^2 + c x + e y + f = 0 $
The equation represents an ellipse if:
$ b^2 - 4 a c < 0 $
or similarly,
$ 4 a c - b^2 > 0 $
The coefficient normalizing factor is given by:
$ q = 64 {{f (4 a c - b^2) - a e^2 + b d e - c d^2} \over {(4ac - b^2)^2}} $
The distance between center and focal point (either of the two) is given by:
$ s = {1 \over 4} \sqrt { |q| \sqrt { b^2 + (a - c)^2 }} $
The semi-major axis length is given by:
$ r_\max = {1 \over 8} \sqrt { 2 |q| {\sqrt{b^2 + (a - c)^2} - 2 q (a + c) }} $
The semi-minor axis length is given by:
$ r_\min = \sqrt {{r_\max}^2 - s^2} $
The latus rectum is given by:
$ l = 2 {{ {r_\min}^2 } \over {r_\max}} $
The eccentricity is given by:
$ g = {{s} \over {r_\max}} $
The distance between center and closest directix point (either of the two) is given by:
$ h = {{{r_\max}^2} \over {s}} $
The center of the ellipse is given by:
$ x_\Delta = { b e - 2 c d \over 4 a c - b^2} $
$ y_\Delta = { b d - 2 a e \over 4 a c - b^2} $
The top-most point on the ellipse is given by:
$ y_T = y_\Delta + {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_T = {{-b y_T - d} \over {2 a}} $
The bottom-most point on the ellipse is given by:
$ y_B = y_\Delta - {\sqrt {(2 b d - 4 a e)^2 + 4(4 a c - b^2)(d^2 - 4 a f)} \over {2(4 a c - b^2)}} $
$ x_B = {{-b y_B - d} \over {2 a}} $
The left-most point on the ellipse is given by:
$ x_L = x_\Delta - {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_L = {{-b x_L - e} \over {2 c}} $
The right-most point on the ellipse is given by:
$ x_R = x_\Delta + {\sqrt {(2 b e - 4 c d)^2 + 4(4 a c - b^2)(e^2 - 4 c f)} \over {2(4 a c - b^2)}} $
$ y_R = {{-b x_R - e} \over {2 c}} $
The angle between x-axis and major axis is given by:
if $ (q a - q c = 0) $ and $ (q b = 0) $ then $ \theta = 0 $
if $ (q a - q c = 0) $ and $ (q b > 0) $ then $ \theta = {1 \over 4} \pi $
if $ (q a - q c = 0) $ and $ (q b < 0) $ then $ \theta = {3 \over 4} \pi $
if $ (q a - q c > 0) $ and $ (q b >= 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} $
if $ (q a - q c > 0) $ and $ (q b < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {\pi} $
if $ (q a - q c < 0) $ then $ \theta = {1 \over 2} {atan ({b \over {a - c}})} + {1 \over 2}{\pi} $
The focal points are given by:
$ F_{1,x} = x_\Delta - s \ cos (\theta) $
$ F_{1,y} = y_\Delta - s \ sin (\theta)) $
$ F_{2,x} = x_\Delta + s \ cos (\theta) $
$ F_{2,y} = y_\Delta + s \ sin (\theta)) $
(I tried to enter the equations without error. If you find an error, please post a comment)
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 $b = a \sqrt{e^2-1}$. $\square$
Best Answer
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.)