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]

Thanks. I have not understood this code yet, but why a peak at 0.25 when the mean value for normal distribution is zero ??

I think what you've given the OP is the distribution of the "distance" from the origin and I don't think that's what's being asked. 0.25 is approximately where the PDF of the distance has a maximum (although I'd say it's closer to 0.3 for the example you gave):

dist = TransformedDistribution[Sqrt[x^2 + y^2], {x \[Distributed] NormalDistribution[0, 0.3], 
    y \[Distributed] NormalDistribution[0, 0.3]}];
d = RandomVariate[dist, 1000000];
skd = SmoothKernelDistribution[d];
Plot[PDF[skd, x], {x, 0, 1.2}]

FindMaximum[PDF[skd, x], {{x, 0.25}}]
(* {2.01764, {x -> 0.301453}} *)

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.

What about using histogram !!

![data = RandomVariate\[NormalDistribution\[0, 5\], 10000\];
Show\[
 Histogram\[data, 20, "ProbabilityDensity"\],
 Plot\[PDF\[NormalDistribution\[0, 5\], x\], {x, -15, 15}, 
  PlotStyle -> Thick\]\]][1]

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];

It certainly does not. Why are you saying that? If you are referring to the figure

that is the distribution of the distance sample points are from the origin. What

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

gives you is a set of points centered on the origin:

The function to obtain random samples from a specified distribution is RandomVariate. Hans used RandomReal. But while that appears to work, the documentation does not mention that RandomReal should work that way.

Got it. FWHM = 2.35 * standard deviation. Thanks for you help.

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

FWHM - Is a measure of the width of the peak. abbreviation of Full width at half maxima. Trying to figure out how the "standard deviation" is related to FWHM !!

or something like this?

xx = RandomReal[NormalDistribution[0, 1], 500];
yy = RandomReal[NormalDistribution[0, 1], 500];
pp1 = Transpose[{xx, yy}];
xx = RandomReal[NormalDistribution[0, .1], 500];
yy = RandomReal[NormalDistribution[0, .1], 500];
pp2 = Transpose[{xx, yy}];
ListPlot[{pp1, pp2}, AspectRatio -> Automatic, 
 PlotRange -> {{-1, 1}, {-1, 1}}, PlotStyle -> {Red, Blue}]