If a circling airplane (AC-130) fires a bullet how does the plane's trajectory alter the shell's flight path?
The bullet leaves the barrel with the combined velocity of barrel muzzle velocity plus the instantaneous aircraft true air speed velocity (actually, the instantaneous true air speed velocity of the tip of the muzzle). That puts the velocity vector of the bullet ahead of the line of the barrel (ahead of the 90 degree wing line). The fact that the barrel is rotating in inertial space is irrelevant. Whether the aircraft is straight and level, turning towards the tank, or away from it, if, at the moment the bullet leaves the muzzle, it is traveling in inertial space towards the tank, (and sufficiently above it for gravity drop) it will hit it.   

And since the bullet's velocity leaving the barrel is a simple vector sum of the muzzle velocity, and the aircraft's instantaneous velocity at that instant. The velocity vector of the aircraft before, (or after), the bullet leaves the barrel is of no relevance.

So if the aircraft is moving due North at 100m/s and the muzzle velocity is 1000m/s due East relative to the aircraft, the bullet is actually traveling 
SQR(100^2 + 1000^2) = 1004.98 ~1005m/s in a direction slightly North of due east by an amount equal to the arctangent of 100/1000 or ArcTan(0.1) = 6.345 degrees
So the bullet is traveling 1004 m/s on a heading of 90-6.34 degrees = 83.66 degrees true. 
So, to hit the tank, the pilot has to fly so that the gun barrel is pointed behind the tank by 6.34 degrees (and above it by a sufficient amount to account for gravity drop)