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

$${x^2+y^2\over c^2} = z^2$$ (1)

$$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = r^2.$$ (2)

Combining (1) and (2) gives

$$(x-x_0)^2+(y-y_0)^2+{x^2+y^2\over c^2}-{2z_0\over c}\sqrt{x^2+y^2}+{z_0}^2=r^2$$ (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

$$(z-z_0)^2 = {x^2+y^2\over c^2}$$ (5)

$$x^2+y^2+z^2 = r^2.$$ (6)

Combining (5) and (6) gives

$$c^2(z-z_0)^2+z2=r^2$$ (7)

$$c^2(z^2-2z_0z+{z_0}^2)+z^2=r^2$$ (8)

$$z^2(c^2+1)-2c^2z_0z+({z_0}^2c^2-r^2)=0.$$ (9)

Using the Quadratic Equation gives

$$z = {2c^2z_0\pm \sqrt{4c^4{z_0}^2-4(c^2+1)({z_0}^2c^2-r^2)}\over 2(c^2+1)}$$

$$= {c^2z_0\pm\sqrt{c^2(r^2-{z_0}^2)+r^2}\over c^2+1}.$$ (10)

So the curve of intersection is planar. Plugging (10) into (5) shows that the curve is actually a Circle, with Radius given by

$$a=\sqrt{r^2-z^2}.$$ (11)