Doubts about direction of normal contact force acting on a tilted rod supported by another body I'm confused about the direction of the normal contact force acting on a body supported by another body. I have read the following :

If a body is being supported by a surface, the normal force acting on the body is always perpendicular to the surface, irrespective of the tilt of the body.
https://physics.stackexchange.com/a/366717/326319

and I'm trying to apply it to different scenarios.

Assume the surfaces are frictionless.
Case 1 : Circular motion of a particle inside a circle in a vertical plane
Reaction force is always directed towards center of circle. This is because of circle properties : line joining particle and center is perpendicular to tangent at particle.
Case 2 : A rod AB rests in equilibrium inside a hemisphere in a vertical plane. Part of the rod is outside the hemisphere.
At B, the reaction force acts towards center. However, reaction force at A is not towards center of circle. Why is this so? The surface supporting the rod is vertical so shouldn't the normal contact force be horizontal ? How are points A and B different?
Case 3 : A rod AB rests in equilibrium inside a sphere in a vertical plane. Whole rod is inside sphere.
Based on case 2, am I right to think that the reaction forces at both A and B act towards center?
Case 4 : A horizontal rod AB rests in equilibrium inside a sphere in a vertical plane. Whole rod is inside sphere.

The reaction forces at A and B are also towards center right?
Update
After doing some research, I found this :

The normal is at right angles to the contact plane


*

*If either surface is differentiable at the point of contact it has a tangent plane there

*If both surfaces have tangent at the contact point then they are necessarily the same plane, or the bodies would intersect.

*If only one has a tangent plane (as in the picture) , use that.

*If neither has a tangent plane there, the direction of the normal is indeterminate.

Source : https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/#toggle-id-1

I also read this discussion.
So I guess, the normal contact force is always perpendicular to the rod if the rod rests on an edge, except at the tip of the rod. My assumptions for the previous cases seem to be correct.
I also concluded the following for the normal contact force acting on a rod in equilibrium against a wall :

There's friction force as well, but I have excluded it.
But what if the edge of the ladder and the edge of the wall intersect at A as shown below?

 A: The “normal” force is defined as the component of the contact force from one object on another which is perpendicular to the surface.  In the first three of the situations you present, there is also a component of the contact force (friction) which is parallel to the surface. This assumes that at least one of the surfaces can be considered locally “flat” at the point of contact.
A: Geometrically speaking, the direction normal to a surface doesn't always exist. When a surface has a sharp angle, it has no defined normal on that angle.
In your case 2, the normal direction is undefined on $A$. That doesn't mean that the contact force doesn't exist: you can for example assume that the system is at rest, and you'll deduce the contact force on $A$ from the rest.
Same on case 5. The best you can do is to develop the contact force on the horizontal and vertical directions, but without knowing beforehand its overall direction. So if the answer in this book assumes that its normal to the rod without any proof, then the author was careless.
Also, as pointed out on another comment above, if there's friction, the contact force is no longer normal (its tangential component is the friction force).
