@Jim Baldwin

Indeed, he didn't ask for that. But I asked myself, whether pp2 given above follows a normal distribution. Not being a specialist in statistics I had no other idea as to check whether the densitiy follows an Exp[ - a x^2 ] - law, which I think it does.

You do get a peak of the density of points at the origin. The density of points for a bivariate normal distribution with mean vector (0,0) does have a peak at (0,0).

Hans introduced the distribution of the "distance" from the origin which doesn't seem to have anything to do with your question. The peak around 0.3 mentioned by Hans is the peak of the probability density function of the "distance from (0,0)" which is a univariate distribution. And that, again, is an interesting distribution but does not appear to me to something that you were asking about.

So no need to modify the Normal distribution. You can get a set of data points with the origin at (0,0) (and therefore the highest density of sample points near the origin) from two independent normal random variables in several ways. Here are 2:

sd = 5/(2 Sqrt[2 Log[2]]);
data = RandomVariate[NormalDistribution[0, sd], {500, 2}];
data = RandomVariate[BinormalDistribution[{0, 0}, {sd, sd}, 0], 500];

OP is for "original poster". In other words, a reference to the person who asked the question or to the question being asked.

Your initial response did contain the what was asked for: how to generate the sample points. Your other response described the distribution of the distance of those points from (0,0) and I don't see that the OP asked for that.

rho = 0.8;  (* Correlation coefficient *)
FWHM = 5;  (* Your desired value for FWHM *)
sigma = FWHM/(2 Sqrt[2 Log[2]]);  (* Convert to equivalent standard deviation for a Normal *)
n = 1000;  (* Sample size *)
xy = RandomVariate[BinormalDistribution[{0, 0}, {sigma, sigma}, rho], n];
ListPlot[xy]

True !! I have to relate a calculation with experimental data, where peak is characterized by FWHM. Thanks by the way.

Here is another piece of code "to analyse" the density of the points obtained, using the assumption that the density is something like Exp[ - a x^2 ]

xx = RandomReal[NormalDistribution[0, .3], 500];
yy = RandomReal[NormalDistribution[0, .3], 500];
pp2 = Transpose[{xx, yy}];
ListPlot[pp2, AspectRatio -> Automatic, 
 PlotRange -> {{-1, 1}, {-1, 1}}, PlotStyle -> {Red, Blue}]
total = Length [pp2]; pts = 
 Table[{d + .05, 
   N@Length[Select[pp2, d < Sqrt[#.#] < d + .1 &]]/total}, {d, 0, 
   2, .1}]
fit = NonlinearModelFit[pts, 
   2 Pi x  b  Exp[-a x^2], {{b, .21}, {a, 6.2},}, x];
ffit = Normal[fit]
Plot[ffit, {x, 0, 2}, PlotRange -> {0, .3}, Epilog -> Point /@ pts]

Run this code several times and look at the plots

Thanks.

In NormalDistribution[0, 1], first variable is the position of the Gaussian peak, defined as mean in Mathematica. However, I am not able to figure out the fact that how the second variable, which is defined as "deviation" in Mathematica, is related to FWHM ? Can you please elaborate that how the number "1" chosen by you is equivalent to the FWHM of 5.

I am not sure whether this does the job

xx = RandomReal[NormalDistribution[0, 1], 500];
yy = RandomReal[NormalDistribution[0, 1], 500];
pp = Transpose[{xx, yy}];
ListPlot[pp, AspectRatio -> Automatic]

but you could have a look at it.