If a photon were exactly at the event horizon of a black hole...?

Absolutely not. There is no circular orbit at, or anywhere near, the event horizon. Light orbits the black hole at the *photon sphere*, NOT the event horizon. The photon sphere is at r=3/2 r_s, where r_s is the Schwarzschild (event horizon) radius. Anywhere inside of r=3/2r_s, a photon with initial motion perpendicular to the hole is captured by the hole.

If a black hole could exist the radius of the event horizon would depend on it's mass. A minimum size black hole would have a diameter of about 3 km. and if it could be seen you would have to be dangerously close as it would have a mass of about 2 of our suns.. The quantum effect would mandate that the thickness of the one way membrane could be no more than one-tenth of a mm thick,no matter the radius of the entity. Unless a photon was emitted from the surface the black hole would remain invisible.

No. The Event Horizon is the maximum distance from a Singularity from which light cannot escape. There is no "zero" or balance point. You are either at or inside the maximum distance from the Singularity from which light cannot escape, or you are outside of that distance.

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Deys a major event on da horizon every time I's near a black hole. I soon in da black hole. Word.

Hypothetically yes, but in reality no. Something will come along and smack the photon below the event horizon.

It really is dependent on conservation of momentum (or equivalently, translational invariance) and the speed of the photon's angular momentum spin.

y (Evt Hzn) = 0x+5 and x+0y = -0 then yes.. it would stay where it was because of the sustaining mass/G2

I think.. :) (best guess)