circlesgeometry
I'm an openGL programmer, trying to construct a sphere in specific circumstances.
Imagine a sphere sliced up into numerous flat horizontal circles.
Given a position along the vertical (Y) axis of the sphere and the radius of the sphere, how do I get the radius of the circle that is on the (X,Z) plane that intersects the Y axis at that point?
Introduce system of coordinates $xy$ plane of which lies on a given plane that crosses a sphere. Equation of the plane is $z = 0$ then, and equation of the sphere $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = R^2$, so after substitution you'll find that $$ (x-x_0)^2+(y-y_0)^2 = R^2-z_0^2 $$ Also you can see that not every plane can intersect a sphere ($R < z_0$) or it might be only 1 point ($R = z_0$, then the plane is a tangent plane), and it might be a circle (if $R > z_0$) with the radius $r = \sqrt{R^2 - z_0^2}$
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
A sphere of radius $R$ is the solution set of $x^2 + y^2 + z^2 = R^2$ in which $x, y,$ and $z$ all vary. A slice of the sphere perpendicular to the $y$-axis has the same equation except that $y$ is a fixed number. So we have $x^2 + z^2 = R^2 - y^2$. With $y$ fixed, this is the equation of a circle with radius $\sqrt{R^2 - y^2}$.