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It is known that the eccentricity of a hyperbola is >1, and that of a circle is 0. These are both conic sections of (the lower nappe of) a (double) cone, with r the radius and h the height of the cone, and theta the angle of the intersecting plane to the vertical axis of the cone:

enter image description here (from wikipedia)

Assume the circle is a horizontal section of a plane with this theoretical cone, and the hyperbola a vertical section of a plane with the same cone. So these 2 planes are perpendicular, and assume these 2 planes intersect right in the point where the circle "touches" the outside of the cone (intersecting conics). In other words: the maximum of the (bottom part) of the vertical hyperbola touches the "left" side of the circle in 1 point. So it is a vertical hyperbola: both foci and midpoint lie on a vertical line (parallel to the Y-axis in a 2-dimensional plane).

enter image description here

Quick version: horizontal circle with unknown radius touches vertical hyperbola with known equation in a 3-dimensional space: can I know the radius of the circle using conic sections and eccentricity? Or some other way?

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Viewing a vertical hyperbola in a direction perpendicular to the cutting plane, the "edges" of the cone appear as the hyperbola's asymptotes.

enter image description here

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 $b = a \sqrt{e^2-1}$. $\square$

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With the origin at the apex, the cone has the equation

$$\frac{z^2}{h^2}=\frac{x^2+y^2}{r^2}.$$

If you slice it with the vertical plane $y=a$, the hyperbola is

$$\frac{z^2}{h^2}=\frac{x^2+a^2}{r^2}$$ and the vertex is found at $x=0$,

$$z_{\text{vertex}}=\frac{ah}r.$$

For this point, the radius is $a$.

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