Equation of cone using sphere and vertex outside sphere

Question

Given equation of sphere and a vertex (v) outside sphere,find equation of cone touching the sphere with vertex (v). note:equation of sphere and vertex is in terms of x,y,z.

Answer 1

Given a sphere $S$ with center at $p_0$ and radius $r$ and a point $p_1$ as vertex outside the sphere, the cone touching the sphere can be obtained using a pencil of lines passing by $p_1$ and tangent to $S$ so

$$ S\to ||p-p_0|| = r\\ L\to p = p_1 + \lambda \vec v $$

where $p = (x,y,z)$

are the elements. Now $S \cap L $ gives

$$ ||p_1-p_0+\lambda \vec v||^2=r^2 $$

or

$$ ||p_1-p_0||^2+2\lambda < p_1-p_0,\vec v > + \lambda^2||\vec v||^2 = r^2 $$

now solving for $\lambda$

$$ \lambda = \frac{-2<p_1-p_0,\vec v > \pm \sqrt{4<p_1-p_0,\vec v>^2-4||\vec v||^2(||p_1-p_0||^2-r^2)}}{2||\vec v ||^2} $$

but the lines should be tangent hence

$$ 4<p_1-p_0,\vec v>^2-4||\vec v||^2(||p_1-p_0||^2-r^2)=0 $$

This is the cone parameterized by $\vec v$. Putting $\vec v = p-p_1$ we obtain the cone equation.

As an example, assuming

$$ p_0 = (0,0,0)\\ p_1 = (5,5,5)\\ r = 2 $$

we have the cone

$$ 23 x^2-25 x y-25 x z+20 x+23 y^2-25 y z+20 y+23 z^2+20 z-150 = 0 $$