How would an event appear to an observer circling it near the speed of light? An observer is circling $\text{Point A}$ once per second with a tangential velocity arbitrarily close to the speed of light $\left(v \lesssim c\right).$
When an event occurs at $\text{Point A},$ what would the observer observe?
 A: As noted above, you need to replace $v=c$ with $v<c$ for this to make sense.
Of course Bob is always traveling with speed $v$, so Alice's clocks always run slow by a factor of $\sqrt{1-v^2}$ in his instantaneous frame (and his run slow in her frame by the same factor).  
But you asked what Bob actually sees.
So let Bob travel around the unit circle (parameterized by arclength) at speed $v$ while Alice stays put at the origin.  Alice says that Bob is at point $0$ at time $0$ (call this event $E$) and at point $vT$ at time $T$ (call this event $F$).  Because his speed is always $v$, she must see his clock advance by $T\sqrt{1-v^2}$ from event $A$ to event $B$, and of course Bob must agree with this.  
If that went by too fast, you can calculate directly that the time that passes on Bob's clock must be
$$\int_0^T \sqrt{1-v^2\cos(vt)^2-v^2\sin(vt)^2}dt=T\sqrt{1-v^2}$$
At event $A$, Bob  receives a light signal that Alice sent (according to her) at time $-1$.  At event $B$, he receives a light signal that Alice sent (according to her) at time $T-1$.  That is, he has seen Alice's clock advance by $T$ while his own clock has advanced by $T\sqrt{1-v^2}$.    Thus he sees her clock moving fast by the same factor that she sees his clock running slow.
A: First, no object, or even elementary particle can move at the speed of light when measured locally.
If you would like to imagine what a photon or massless particle (that moves with speed c when measured locally) would observe as it circles around the center, you would have to use an inertial frame for the photon.
Now massless particles, and the photon do not have a frame of reference. When you try to do the math for the photon, trying to calculate things from its frame, you will get divisions by 0.
Now if you would like to still think about what an object near that speed can observe, you should use a neutrino.
Your question, which you should edit, will be "a neutrino, as an observer is moving in circular motion around a center".
Now in this case, the neutrino does have a reference frame, and you can do the math. You defined the radius as 1 lightsecond. The photon will take this path in a second. So anything you would observe, would reach the observer (the neutrino) in 1 second.
Now because of special relativity, the photons that will come from the center towards the neutrino, will seem to be traveling still at speed c when viewed from the neutrino's frame.
The observer could still see the event, because information from the event would still reach the observer, but because the relativistic angular speeds of the observer, it would see the event in fast motion (and from a rotating angle, where the angle of view would rotate with speed almost c).
Let's say there is a clock in the center. The observer (neutrino) would see the clock in the center ticking faster (then its own clock) as per SR. Because as per SR, the observer, who is moving at relativistic speeds, will see the event (at rest in the center) in a faster pace (then it would see it if the neutrino would move slow).
As per SR there is relativity of simultaneity, that means that if there would be another observer in the center, that observer would say the same thing about the neutrino, because relativity is symmetric. Now that is only true if the two observers are moving in a linear motion, non-accelerating compared to each other.
But in your case, the neutrino is constantly accelerating (angular momentum), and that is absolute. The observer in the center would see the neutrino move fast, but the other observer with the neutrino would see the event in the center rotating (and not moving in a linear motion).
In this case, SR would not give you a solution. Though, I think that the observer in the center would see the neutrino's clock tick slow (compared to the clock in the center). And the observer co-moving with the neutrino would see the clock in the center tick fast (compared to the clock at the neutrino).
