Normal-like distribution over a bounded area Is there a distribution that resembles the gaussian (normal) distribution, but such that it's probability density is nonzero only over a defined segment.
The question emerged when I tried to model the 'bullet spread' within a circle. Gaussian distribution works fine, but there is always a chance that the bullet would hit outside the circle. So I'd like to find a distribution very similar to Gaussian, but with property that the probability outside the defined segment (or circle) is zero.
EDIT: Yes, actually I mean a disk, not a circle.
EDIT: And yes, I need only a one-dimensional distribution (along the radius of a disk) which will be circular-symmetrical (not dependent on angle).
 A: The VonMises distribution is similar to the normal, but is used with circular data and is defined just on the interval of a circle (0-360 degrees, or 0-2pi radians).  
The Beta distribution is defined from 0 to 1 (but could be scaled to other intervals), with the parameters equal it is symmetric and for many values bell shaped.
A: This is an old question, but it's still relevant for new readers.  I'm surprised that nobody mentioned the Raised Cosine distribution.
With mean $\mu$ and spread parameter $s$ it is perfectly bounded into  $[\mu - s, \mu + s]$  and its probability density function (PDF) has a bell shaped curve as well.
A: You could use a truncated normal distribution.  It's just a normal distribution that you only consider an interval for.  You need to rescale it to make sure that the pdf integrates to 1.  But this sounds to me to be exactly what you're looking for.
A: It seems what is searched for is a uniform distribution on a disk, which I will take to be (the interior of) the unit circle.  We can parametrize by $(r,\theta)$ so we have $0 \le r \le 1$ and $0 \le \theta \le 2\pi$.  We can let $\theta$ have a uniform distribution, independent of $R$, and must find the distribution of $R$ that gives a uniform distribution on the circle.  Since probability must be proportional to area, we have for $0 \le a \le b \le 1$ that 
$$\DeclareMathOperator{\P}{\mathbb{P}}
   \P(a\le R \le b) \propto \pi b^2 - \pi a^2
$$
and taking $a=0$, $b=1$ gives $F_R(r)= r^2$.  Then the density is the derivative $f_R(r)=2r$. The joint density of $R$ and $\theta$ then becomes
$$
  f(r,\theta)=\frac1{2\pi}\cdot 2r = \frac{r}{\pi}
$$
This is easy to simulate from, the sum of two independent uniforms have a triangular (and symmetric) distribution, sometimes described as a "tent" distribution.  We only want the left part of the tent, which we can get by mirroring the distribution in a vertical line at the top (mode) of the tent.  Simulating this in R gives:

The R code for the simulation is:
set.seed(7*11*13)
rleft_tri  <-  function(n) {
    T  <-  runif(n)+runif(n)
    val  <-  ifelse(T <= 1,T, 2-T)
    val
}

rdisk  <-  function(n)  {
    val  <-  cbind(  rleft_tri(n),  2*pi*runif(n) )
    colnames(val)  <-  c("R","Theta")
    val
    }

#

library(plotrix)
par(bg="antiquewhite")
points  <- rdisk(10000)         plot(c(-1,1),c(-1,1),type="n",axes=FALSE,xlab="",ylab="",xlim=c(-1.1,1.1),ylim=c(-1.1,1.1))
    draw.circle(x=c(0,0),y=c(0,0),radius=1,col="aquamarine")
    points(with(as.data.frame(points),cbind(R*cos(Theta), R*sin(Theta))),pch=".",col="red",cex=2)

Note that this is a special case of @Greg Snow's old answer, as the "left tent" distribution is a beta distribution with parameters $a=2, b=1$.  But the above code for simulating it is probably faster than general code for simulating from a beta (or would be so if programmed in C).
A: +1 for the rejection-sampling answer.
Could you also sample from the Beta distribution where $\alpha$ (aka shape1) is 1 and $\beta \gt 1$ (aka shape2)? This is defined on [0,1], so multiply by the radius of the disc, and you'll have zero probability of selecting points at the radius or greater.
The upsides include: a) there's a zero probability of selecting a distance greater than or equal to the radius, and b) you can do straightforward sampling rather than things like rejection sampling.
The downsides include: a) it's fidgety close to 0 and b) the distribution is not "very similar" to the Gaussian. (It's much more peaked near 0 -- i.e. in the center -- than the Gaussian, though that might indeed be what the OP wants.)
