 Normals through a point on a ring torus

For the surface of revolution of the circle with center $(a,R)$ and radius $r<R$ in the upper half-plane, use $X(u,v)=(a+r\cos u,(R+r\sin u)\cos v,(R+r\sin u)\sin v)$ and unit normal $n=(\cos u,\sin u\cos v,\sin u\sin v)$. Exactly four surface normals pass through the displayed $P$. The equation $P-X=tn$ forces $(t+r)\cos u=0$. The case $t=-r$ would put $P$ at radial distance $R$, so $\cos u=0$; its nonzero radial coordinate then forces $v=0$ or $\pi$. All four combinations work. By contrast, infinitely many normals pass through $Q$: both complete circles with $u=\pi /2$ or $3\pi /2$ have radial normals through that point. This is a concrete illustration of degenerate distance-critical sets versus isolated critical points.