Is the speed of light constant in all directions? Consider the shape of the waveform emitted by a moving source:

(with the horizontal axis "x" and the vertical axis "y").
According to the stationary observer, the wavefront of a short pulse emitted like this will reach a given distance along y before it will along x.
Does the speed of light depend on direction? 
EDIT: After researching the answers I personally found it easiest to understand by looking at the emission of the first pulse in this video. The picture above is misleading because it makes you think the red dot is the original center of each wave.
 A: Actually the answer is "we don't know", because we have never measured speed of light in only one direction. All experiments measure "there and back again" speed.
Speed of light being the same in all directions is an axiom postulated by Einstein. That also means that you can't use SR/GR to prove/disprove it (see Kuhn Cycle) because c being constant is foundational for the current paradigm.
P.S. as mentioned by others, the Doppler shift does not affect the speed of light.
Update: there is a video: https://www.youtube.com/watch?v=pTn6Ewhb27k
A: The observed wavelength does change, and this is called the doppler effect. But the speed does not change. The statement "...the wavefront of a short pulse emitted like this will reach a given distance along y before it will along x" does not follow from any logical reasoning. Following the same logic you will conclude that when the source is next to the observer the speed of light will be infinite (because the next front will reach you immediately).
What actually happens is that 
the front of the peak was emitted when the source was at the center of the smaller circle,  so assuming that c is constant in both directions will allow you only to conclude that such front will reach the same distance in both directions at the same time. The successive fronts were emitted at different distances from the observer, thus the packing up of wavefronts towards the observer.     
A: According to all observational evidence (including the original Michelson-Morley experiment) speed of light is constant in all directions. The confusion comes from misinterpretation of the picture you attach. I propose to understand it as follows.
Say, you have a lightsource that emits pulses at a given frequency. Each pulse propagates at a constant speed in all directions forming a wavefront that is shown as a blue circle. Then, if the lightsource is moving, the center of each following wavefront circle will be shifting, exactly as you see on the picture. As the time goes, each circle evenly expands while the source emits new pulse. 
The picture you show is most probably an illustration of the Doppler shift, like this one:

(from http://www.radartutorial.eu/11.coherent/co06.en.html)
All circles expand with a constant speed here.
A: Maxwell's equations, when stated in the following form:
$$
 = ∇×, \hspace 1em  = -∇φ - \frac{∂}{∂t}, \\
∇· = 0, \hspace 1em ∇× + \frac{∂}{∂t} = , \\
∇· = ρ, \hspace 1em ∇× - \frac{∂}{∂t} = , \\
∇· + \frac{∂ρ}{∂t} = 0
$$
transcend all assumptions about light speed (as it makes no direct reference to light speed), and about causal structure, itself, transcending the distinction between relativity and non-relativistic theory. The place where the question and assumptions appear lies not with these equations, but in the constitutive relations, which link the fields $(,)$ and $(,)$ - and this has changed over time.
Maxwell's actual theory, when restricted to the case of isotropic media, posed constitutive laws that in today's language would be written as:
$$ = ε( + ×), \hspace 1em  = μ( - ×).$$
The inclusion of the $-×$ is actually posthumous (an oversight by Maxwell, corrected by Thomson). When the constitutive relations are expressed with $ ≠ $ they are what was referred to at the time as the "moving" form of Maxwell's equations, while the form with $ = $ was referred to as the "stationary" form of Maxwell's equations. (Also: Maxwell pushed the $×$ term into the $$ versus $(φ,)$ relation, by writing it - instead - as $ = -∇φ - ∂/∂t + ×$, where we would today define $$ as just including the first two terms; but the end result is equivalent).
A reference to light speed motion appears in the equations, with the speed given by $V = 1/\sqrt{εμ}$: the wave speed for the field. The equations respect the Galilei transform group - they are non-relativistic. Maxwell took pains to show the Galilei-covariance in his treatise, though he mangled the math (which contributed to the confusion by his successors and their attempted treatments of the theory, after his demise.)
Because of the appearance of $$, an outward-directed wave will appear as a sphere expanding at speed $V$ with a center that drifts in a way noted by the velocity $$; so the speed you measure in different directions will be different. That's the origin of the idea, and the whole point of the Michelson-Morley experiments was to find an in vacuo measure of $$. I emphasis the "in vacuo" part, by the way, because there is always a drift speed, if you're talking about propagation in a medium, relativity or not. It's not a matter of if, but of degree. More on this below.
One of the main objections Einstein had to this is that the appearance of inequivalent forms premised on some kind of background velocity made a distinction between different moving frames of reference that did not accord with what we observed in nature. In fact, it was common practice at the time (even before the Michelson-Morley experiments) to just use the stationary form of Maxwell's constitutive relations - at least for celestial observations where light is propagating in a near-vacuum.
At the time, Einstein was young and (clearly) didn't have a full grasp of Maxwell's theory (at least not in any version that Maxwell, himself, published); he was using a hand-me-down version of electrodynamic theory that came from Hertz, which was one of the cases in point of the "confusion postdating Maxwell" I was referring to. So he made no direct reference to $$ itself in his 1905 paper, only an oblique reference to it, stating that (with his new formalism) it is now superfluous. But refer to it he did: the question of what the "moving form" of Maxwell's equations ought to be was the whole point of the title "On the Electrodynamics of Moving Bodies", and is why it had that name. It would have been more accurate to have titled the paper "Why Maxwell Is Always Stationary" or "Why $ = $, Always".
In today's language, Einstein asserted that the following constitutive relations should hold in the vacuum:
$$ = ε, \hspace 1em  = μ,$$
irrespective of the motion of the observer. It's not equivalent of Maxwell's version, since these do not respect the Galilei transforms; but (instead) the Lorentz transforms.
As a footnote, Lorentz also posed a formalism for Maxwell's theory which introduced the Lorentz transforms. However, it was still non-relativistic: the constitutive laws missed the key term that actually distinguishes between the relativistic and non-relativistic versions. Einstein, himself, pointed out this discrepancy later on (I think in 1920).
The two versions of the constitutive laws can be combined in a unified framework that sets in clear relief who is who, what and where.
Essentially, as part of the exercise of reconciling the old pre-relativistic version of Maxwell's theory with the version emerging out of Einstein's paper, and determining whether there actually was some kind of paradigm break here, Einstein and Laub in 1908 presented a version of the constitutive laws that applies to both vacuua and media, while also respecting relativity. At the same time, Minkowski also presented an equivalent formulation of the same in a paper where he first introduced the Minkowski geometry: it's now known as the Maxwell-Minkowski constitutive relations.
In today's language, they would be written as:
$$ + \frac{1}{c^2} × = ε( + ×), \hspace 1em  - \frac{1}{c^2} × = μ( - ×).$$
That's the relativistic version of the older Maxwell theory. Lorentz's papers did not have the $1/c^2$ terms in his version of the constitutive laws, which is why they were actually non-relativistic; so Lorentz was actually equivalent to Maxwell, not to what we today (mis-)label Maxwell. Had those correction terms been present in any of Lorentz' works, it would have been correct to credit him, instead of Einstein, with the discovery of Special Relativity; but he missed the mark.
The equations permit wave motion at a speed $V = 1/\sqrt{εμ}$, as before. If $c ≠ V$, then the above-mentioned "center-drift" will occur for outward light propagation. You may not be able to observe the center-drift directly from the center, itself, but you can certainly observe it from a vantage point off to the side - thus answering the original question.
In a vacuum, if $c = V$, the above equations continue to hold, even with the $$ vector still there! But, you can (almost) work out mathematically that the $$ cancels and the equations reduce to the form $ = ε$, $ = μ$. So, what Einstein said was literally true.
Oh, but I said "almost". There is one exception: if $|| = V$. Then a residual of $$ remains behind, even in a vacuum, even in Relativity, as $V → c$. So, it's not entirely superfluous. I don't know if anyone's ever tested for this. It might be relevant in plasma physics, or in cases where one actually does have some kind of background light-speed medium.
So, when the different versions of the constitutive laws are combined, they generalize to the form:
$$ + α × = ε( + β ×), \hspace 1em  - α × = μ( - β ×).$$
These equations are covariant under transforms that respect the following geometric invariants:
$$
α(dx^2 + dy^2 + dz^2) - β dt^2,\\
dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z} + dt \frac{∂}{∂t}, \\
β \left(\left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2\right) - α \left(\frac{∂}{∂t}\right)^2.
$$
The different cases may be enumerated as follows:

*

*$α = 0$, $β ≠ 0$: non-relativistic theory; where $β$ can be normalized to 1.

*$αβ > 0$: the 3+1-dimensional Minkowski geometry of Special Relativity, with light speed $c = \sqrt{β/α}$. Again, $β$ can be normalized to 1.

*$α ≠ 0$, $β = 0$: the Carroll universe - where $c = 0$. It's named that, because in it, things move without going anywhere and 0 is both an absolute speed and speed limit. Everything that moves at all is a tachyon.

*$αβ < 0$: a 4-dimensional Euclidean geometry of a timeless space: where time is a spatial dimension, and there is no time dimension at all.

*$α = 0$, $β = 0$: the Static universe - not in the sense of Cosmology (that's a different usage of the term "Static universe"), but in the sense of kinematic symmetry groups; it's associated with the "static group"; and in this geometry, all speeds are absolute, not just 0 (in Carroll), c (in Relativity) or infinity (in non-relativistic theory). For the static universe, the parts delimited by $α$ and $β$ are separately invariants, instead.

And: expanding on what has already been noted above, the $$ term becomes superfluous (just as Einstein said it would), precisely when $βεμ = α$.
So the question of direction-dependence and center-drift in a vacuum is not a matter of perspective at all. As already noted: you can tell whether there is a center drift or not by looking at it from a vantage off to the side. Rather, the question is the one that runs central to the very question: $α = 0$ & $β ≠ 0$ or $αβ > 0$?
