Find the set of points on the tangent lines drawn from O to…

Find the set of points on the tangent lines drawn from $O$ to the sphere $x^2+y^2+z^2-10x+2y+10=0$.

Explanation of Method

All tangent lines from the origin to a sphere form a quadric cone.

Step 1: Sphere

\[ (x-5)^2+(y+1)^2+z^2=16,\qquad C(5,-1,0),\ R=4,\qquad |OC|^2=26\gt 16 . \]

So $O$ is outside the sphere, and each tangent segment has length $\sqrt{26-16}=\sqrt{10}$.

Step 2: Tangency

The line $OM$ is tangent if and only if $C$ is at distance $4$ from it: \[\frac{|\overrightarrow{OC}\times\overrightarrow{OM}|^2}{|OM|^2}=16,\qquad \overrightarrow{OC}\times\overrightarrow{OM}=\{-z,\,-5z,\,x+5y\},\]

\[ 26z^2+(x+5y)^2=16(x^2+y^2+z^2) . \]

\[15x^2-10xy-9y^2-10z^2=0\]

The equation is homogeneous of degree $2$, so it describes a cone with vertex $O$. It touches the sphere along a circle with center $\tfrac{10}{26}\overrightarrow{OC}$ and radius $\tfrac{4\sqrt{10}}{\sqrt{26}}$.

$O$ $C(5,-1,0)$

Tangent lines from $O$ (teal) touch the sphere along the red circle; each tangent segment has length $\sqrt{10}$. Together they form the cone $15x^2-10xy-9y^2-10z^2=0$.

Result

\[\boxed{\,15x^2-10xy-9y^2-10z^2=0\,}\]