Tangential planes on a surface at the points of intersection with the sphere I was looking at a question in an old exam at it happens that someone has already asked it. However, I have one question regarding the formulation of the answer:

Prove that tangent planes to the surface $a(xy+yz+xz)=xyz,a>0$ at the points of intersection with the sphere $x^2+y^2+z^2=r^2$ cut off segments on the coordinate axes whose sum is constant.

In the accepted answer, it is suggested to rewrite the equation of the surface as $\frac1x+\frac1y+\frac1z=\frac1a$ in order to use $f(x,y,z)=\frac1x+\frac1y+\frac1z-\frac1a$ and get $\nabla f(x,y,z)=-\left(\frac1{x^2},\frac1{y^2},\frac1{z^2}\right)$ and the equation of the tangent plane at a point of intersection $(x_0,y_0,z_0)$ with the sphere is: $$\frac1{x_0^2}(x-x_0)+\frac1{y_0^2}(y-y_0)+\frac1{z_0^2}(z-z_0)=0.$$ It follows that the lengths of the segments are $$x=x_0^2\frac{x_0y_0+y_0z_0+x_0z_0}{x_0y_0z_0}\\y=y_0^2\frac{x_0y_0+y_0z_0+x_0z_0}{x_0y_0z_0}\\z=z_0^2\frac{x_0y_0+y_0z_0+x_0z_0}{x_0y_0z_0}$$ and $$\begin{aligned}x+y+z&=(x_0^2+y_0^2+z_0^2)\frac{x_0y_0+y_0z_0+x_0z_0}{x_0y_0z_0}\\&=\frac{r^2}a\end{aligned}$$
However, there are intersection points on the axes, so, $f(x,y,z)=a(xy+yz+xz)-xyz$ gives $\nabla f(x,y,z)=(ay+az-xz,ax+az-xz,ax+ay-xy).$
If $$(x_0,y_0,z_0)=(\pm r,0,0), \nabla f(x_0,y_0,z_0)=(0,\pm ar,\pm ar),$$ the tangential plane is parallel to the $x$ axis, moreover, the $x-$ axis lies in the plane.

How is then the length of a segment on the axis defined? Should we isolate this as a special case?

 A: tl; dr: Presumably the examiner forgot to stipulate $xyz \neq 0$.

We may partition the solution set $S$ of
$$
a(xy + yz + xz) = xyz
$$
into the set $G$ (of good points) for which $xyz \neq 0$ and the set $B$ (of bad points) where $xyz = 0$.
By the answer to the linked question, dividing the defining equation of $S$ by $xyz$ yields the equation
$$
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{a}
$$
whose solution set is $G$, where the stated conclusion is true:

If $S^{2}(r)$ is the sphere is radius $r$ centered at the origin, then for each point $p$ of $G \cap S^{2}(r)$, the tangent plane to $G$ at $p$ intersects the Cartesian coordinate axes at three points for which the sum-of-distances to the origin is $r^{2}/a$.

This conclusion does not extend to points of $B \cap S^{2}(r)$ regardless of how we attempt to extend definitions. There are at least two ways to see this:

*

*The positions of the points on the sphere and on a coordinate plane do not depend on $a$, so the asserted sum-of-distances $r^{2}/a$, which depends on $a$, cannot be correct without qualification.

*More explicitly, $B$ is the union of the three coordinate planes. A tangent plane to $S$ (i.e., to $B$) at a point of $B \cap S^{2}(r)$ is a coordinate plane, which (instead of intersecting each axis at one point) contains two of the axes and intersects the third at $0$.

