So I need to find an intersection of a sphere and a surface. The equation of the sphere is $x^2+y^2+z^2=r^2$ and surface is $a(xy+yz+xz)=xyz$ where $a \gt 0$.
I am not even sure this is possible, I tried using spherical coordinates. I am solving an old exam and this is the part of a task:
Show that tangent planes in the points where the above surface intersects the above sphere cut off the parts on coordinate axes whose sum is a constant.
Thanks for your help!
Let the surfaces meet at the point $(e, f, g)$. Then the equation of the tangent plane to the surface $x^2+y^2+z^2=r^2$ at this is point is given by
$2e(x-e)+2f(y-f)+2g(z-g)=0$. The $x$ , $y$ and $z$ intersect of this plane with the coordinate axes are
$x=\frac{e^2+f^2+g^2}{e}$, $y=\frac{e^2+f^2+g^2}{f}$, $z=\frac{e^2+f^2+g^2}{g}$.
\begin{array} $x+y+z&=&\displaystyle(e^2+f^2+g^2)\big(\frac{1}{e}+\frac{1}{f}+\frac{1}{g}\big)\\ &=&\displaystyle(e^2+f^2+g^2)\big(\frac{ef+fg+eg}{efg}\big)\\ &=&\displaystyle\frac{r^2}{a}=const. \end{array}
Note that $(e, f, g)$ lies on both surface.
Try find the next tangent plane to the second surface at the point of intersection as proceed as above to get the same result $\frac{r^2}{a}.$
Note: Write the 2nd surface of the form $\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a}$ so that the equation of the required plane at $(e, f, g)$ takes the form $\displaystyle\frac{1}{e^2}(x-e)+\frac{1}{f^2}(x-f)+\frac{1}{g^2}(z-e)=0.$