How to find the tangent cone of the sphere A given sphere:
$$x^2+y^2+z^2+2x-4y+4z-20=0$$
How to find the tangent cone of it ?
the vertex of the cone is $(2,6,10)$
thanks very much.
 A: The sphere center is $(-1,2,-2)$ and radius is $\sqrt{29}$. The distance between the given point $(2,6,10)$ and the center of the sphere is $13.$ The axis of the cone will be the ray from $(2,6,10)$ through the sphere center, and the angle at the top of the cone will be $\arcsin \sqrt{29}/13$. It may be involved to turn this into a three variable equation for the cone, but this is a start.
EDIT: A kind of vector equation for the cone: the vector connecting $(2,6,10)$ to the sphere center has components $(3,4,12)$ (and length 13). So the cone, in vector form, is the set of vectors $(2,6,10)+u$ such that $u \cdot (3,4,12)=13|u|\cos \theta$, and here $\cos \theta = \sqrt{140}/13.$ Then by filling in $u=(x,y,z)$ one gets an equation at least for the vectors $u$ to be added to $(3,4,12)$ to give points on the cone.
NOTE: I just fixed the references to the trig functions of the angle $\theta$ formed by the axis of the cone and its curved side. The relevant triangle has hypotenuse 13 and opposite side $\sqrt{29}$ --- I had things backward before. I also just replaced the description by $(2,6,10)+u$ since the "basepoint" of the vector on the cone should be at the vertex of the cone, which is the given point $(2,6,10)$.
