# 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.

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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 $(2,6,10)$.
Check that: I think the sphere's center is $\,(-1,2,-2)\,$ –  DonAntonio Jan 2 '13 at 19:46
Yes that was a typo. I'll fix. I used the correct center $(-1,2,-2)$ when getting the vector connecting $(2,6,10)$ to the center as $(3,4,12)$ so only the typo needs fixing. Thanks. –  coffeemath Jan 2 '13 at 23:28
thanks, and I don't know why the vector form of cone can be written as $(3,4,12)+u$ such that $u \cdot(3,4,12)=13|u|\cos \theta$,suppose the tangent point is $(a,b,c)$,and $u=(a-2,b-6,c-10)$ satisfy $u \cdot(3,4,12)=13|u|\cos \theta$,but$(3,4,12)+u$ seems not in the cone. maybe I have misunderstood your meaning –  Laura Jan 3 '13 at 4:01
Sorry, another case of going too fast. It should be $(2,6,10)+u$ not what I had, which was $(3,4,12)+u$. The "origin" or base of the vector should be at the vertex of the cone, which is your given point $(2,6,10)$. I'll fix that. –  coffeemath Jan 3 '13 at 4:23