Let S be the center of the sphere $\sigma$ and let N be the foot of a normal to the plane $\pi$ from the point S. Denote r the radius of the sphere $\sigma$ and h = NS the height of the center S above the plane $\pi$. First, assume that $h \ge r$. The tangency circle $(L)$ of the cone with the vertes $M \in \pi$ is the intersection of the sphere $\sigma$ with a sphere $\tau$ centered at the midpoint T of the segment SM and with radius $TS = TM = \dfrac{MS}{2}$. The plane $\lambda$ of the tangency circle (L) intersects the line $m \equiv SM$ at a point L, the center of the circle (L). Since the problem is symmetrical with respect to rotation around the axis $n \equiv NS$, we need to consider only a point M moving on a fixed line $p$ in the plane $\pi$ passing through the point N, find the locus of these tangency circles (L) and rotate the locus around the axis $n \equiv NS$. Let $\mu$ be the plane passing through the points M, N, S. The spheres $\sigma, \tau$ project to the plane $\mu$ as circles (S), (T) intersecting at points $A, B$, the plane $\lambda$ of the tangency circle (L) as a line $l \equiv AB,\ l \perp m$. The center $L \equiv l \cap m$ of the tangency circle (L) is identical with the midpoint of the segment AB. Since MS is a diameter of the circle (T) and the angle $\angle MNS = 90^\circ$ is right, the circle (T) always passes through the point N. From the right angle triangle $\triangle MAS$ with the right angle $\angle MAS = 90^\circ$ and the altitude AL to the hypotenuse MS, we have $AL^2 = LS \cdot LM$ $MS \cdot LS = (LM + LS) \cdot LS = LM \cdot LS + LS^2 = AL^2 + LS^2 = SA^2$ where r = SA is a constant radius of the sphere $\sigma$ and its projection, the circle (S). Let the line $l \equiv AB$ intersect the line $n \equiv NS$ at a point Q. The right angle triangles $\triangle MNS \sim \triangle LQS$ are similar, having the angle $\angle MSM \equiv \angle QSL$ common. Consequently, $\dfrac{QS}{LS} = \dfrac{MS}{NS},\ \ QS = \dfrac{MS \cdot LS}{NS} = \dfrac{AL^2}{NS} = \dfrac{r^2}{h}$ Thus we see that the length of the segment QS does not depend on the position of the point M on the line p. Since the angle $\angle QLS = 90^\circ$ is right, the locus of the points L is a circle with diameter QS. Rotating this circle around the axis $n \equiv NS$, the locus of the centers of all possible tangency circles (L) is a sphere $\omega$ with diameter $QS = \dfrac{r^2}{h} \le r$. If we assume h < r instead, then $QS = \dfrac {r^2}{h}> r$, the locus sphere $\omega$ intersects the given sphere $\sigma$ in a circle (N) lying in the given plane $\pi$ and the locus of all possible tangency circles (L) is only the portion of the locus sphere $\omega$ inside the given sphere $\sigma$.

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Uhm ... isn't this construction simply the inversion of the plane in the sphere??