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Cone-Sphere Intersection

Let a Cone of opening parameter $c$ and vertex at $(0,0,0)$ intersect a Sphere of Radius $r$ centered at $(x_0, y_0, z_0)$, with the Cone oriented such that its axis does not pass through the center of the Sphere. Then the equations of the curve of intersection are

$\displaystyle {x^2+y^2\over c^2}$ $\textstyle =$ $\displaystyle z^2$ (1)
$\displaystyle (x-x_0)^2+(y-y_0)^2+(z-z_0)^2$ $\textstyle =$ $\displaystyle r^2.$ (2)

Combining (1) and (2) gives
\begin{displaymath}
(x-x_0)^2+(y-y_0)^2+{x^2+y^2\over c^2}-{2z_0\over c}\sqrt{x^2+y^2}+{z_0}^2=r^2
\end{displaymath} (3)

$x^2\left({1+{1\over c^2}}\right)-2x_0x+y^2\left({1+{1\over c^2}}\right)-2y_0y$
$+({x_0}^2+{y_0}^2+{z_0}^2-r^2)-{2z_0\over c}\sqrt{x^2+y^2}=0.\quad$ (4)
Therefore, $x$ and $y$ are connected by a complicated Quartic Equation, and $x$, $y$, and $z$ by a Quadratic Equation.


If the Cone-Sphere intersection is on-axis so that a Cone of opening parameter $c$ and vertex at $(0,0,z_0)$ is oriented with its Axis along a radial of the Sphere of radius $r$ centered at $(0,0,0)$, then the equations of the curve of intersection are

$\displaystyle (z-z_0)^2$ $\textstyle =$ $\displaystyle {x^2+y^2\over c^2}$ (5)
$\displaystyle x^2+y^2+z^2$ $\textstyle =$ $\displaystyle r^2.$ (6)

Combining (5) and (6) gives
\begin{displaymath}
c^2(z-z_0)^2+z2=r^2
\end{displaymath} (7)


\begin{displaymath}
c^2(z^2-2z_0z+{z_0}^2)+z^2=r^2
\end{displaymath} (8)


\begin{displaymath}
z^2(c^2+1)-2c^2z_0z+({z_0}^2c^2-r^2)=0.
\end{displaymath} (9)

Using the Quadratic Equation gives
$\displaystyle z$ $\textstyle =$ $\displaystyle {2c^2z_0\pm \sqrt{4c^4{z_0}^2-4(c^2+1)({z_0}^2c^2-r^2)}\over 2(c^2+1)}$  
  $\textstyle =$ $\displaystyle {c^2z_0\pm\sqrt{c^2(r^2-{z_0}^2)+r^2}\over c^2+1}.$ (10)

So the curve of intersection is planar. Plugging (10) into (5) shows that the curve is actually a Circle, with Radius given by
\begin{displaymath}
a=\sqrt{r^2-z^2}.
\end{displaymath} (11)



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© 1996-9 Eric W. Weisstein
1999-05-26