No, it's still just a circle, centered at the point where the source was at T=0. Remember, you're not moving in your own frame of reference. What you observe is the source moving, and at some point (T=0), it emits a flash of light, which expands at C in a sphere that is also at rest. The source continues moving as before, past the center of the sphere but always within it, since obviously it is moving at less than C. Because of this, it should be noted that even if it continues to emit light, continuously or in one or more subsequent flashes, the light that it emits at T>0 will still be behind the T=0 sphere in every direction, because, the source being inside the initial sphere, these new wavefronts will never catch the initial one.

Well, maybe expressed like this :

 

a) in the rest frame of the source of the flash, a circle. We express every "ray" with an angle a

b) in the moving frame, again a circle. let every "ray" be determined by angle b

 

Then, we match, or link, every ray in a with b, and the function b(a) is not the identity, so that in the moving frame, it could be described as a gray-shaded interior disc of radius c.

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Oh, I found a simpler way to express this :

 

- O' moves along x, with speed v towards O, the gamma time-dilatation factor is [math]\gamma>1[/math].

 

-Consider an object moving with speed vy in O

 

=> no space contraction along y'

=> time dilation gamma, this factor acts on every direction, hence

 

=> speed vy'=vy/[math]\gamma[/math]<vy

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Imagine a detector array in O', moving along x to the right with speed v

 

-) This detector does not move towards O'

 

-)Now we consider a light ray emitted in O, in direction y.

 

In this case, On the detector, it's obvious that the trace is not along y, because the decector cell at O (origin of O) changes, and moves to the right with speed v, seen from O.

 

-) in O' we deduce (cehck) that several cells along x are activated, so the light moves not purely along the perpendicular direction.