Tetrahedron from its inscribed sphere

Question

I'm facing a geometrical problem:

Given a sphere $S$, I want to calculate the vertices of the tetrahedron $T$ whose inscribed sphere is $S$. In other words I want to calculate a tetrahedron from its inscribed sphere.

If anyone knows the solution, don't hesitate to share with me.

Thanks in advance

Answer 1

The radius of the circumsphere is three times the radius of the inscibed sphere. Hence one tetrahedron with the sphere of radius $1$ around the origin would be given by the vertices $(0,0,3)$, $(\sqrt 8,0,-1)$, $(-\sqrt 2,\sqrt 6,-1)$, $(-\sqrt 2,-\sqrt 6,-1)$.

Answer 2

If the radius of inscribed sphere is $r$ then the edge length of the tetrahedron is $2r\sqrt{6}$ & radius of circumscribed sphere is $3r$

Then in general form, for inscribed sphere with a radius $r$ centered at the origin, the vertices of the tetrahedron are $(0, 0, 3r)$,$\left(2r\sqrt{2},0, -r\right)$, $\left(-r\sqrt{2},r\sqrt{6}, -r\right)$, $\left(-r\sqrt{2}, -r\sqrt{6},-r\right)$