I'm arriving a bit late to the party, but I have an alternative approach that could be of interest to some readers.
All of the other solutions proposed (as of this writing) use variations on the acceptance/rejection technique of sampling. This works well enough for your specific problem, where the difference in the intensity of the process between the areas inside/outside the circle is comparatively small. However, as others have already pointed out, efficiency can become a problem in the limit of large differences between the "background" and the "features" in your problem -- you could have to generate a large number of "rejects" for each "good" sample.
However, if we think of your problem as an example of a (nonhomogeneous) Poisson spatial point process, we can use the additivity property of the Poisson distribution to assemble a sample from your complicated intensity function by combining realizations of simpler, easier to generate, point processes on the square and the circle in isolation, with little or no "wasted" computational effort.
Before we start, let's build a couple of helper routines to generate single, uniformly distributed, points on general rectangles and circles:
generateUniformDistributedPointOnRectangle[{{left_, bottom_}, {right_, top_}}] :=
{RandomVariate[UniformDistribution[{left, right}]],
RandomVariate[UniformDistribution[{bottom, top}]]}
generateUniformDistributedPointOnCircle[{{centreX_, centreY_}, radius_}] :=
Module[{u, theta, r},
u = Total[RandomVariate[UniformDistribution[{0, radius}], {2}]];
theta = RandomReal[{0, 2 Pi}];
r = If[u > radius, 2 radius - u, u];
{centreX, centreY} + {r Cos[theta], r Sin[theta]}
]
Our plan of attack is now to generate $N_{S}$ samples from the entire square and $N_{C}$ samples from just the circular area such that when we combine the two samples, we are left with $N$ samples from our target distribution. So, how to determine the correct values (and distributions) for $N_{S}$ and $N_{C}$?
Our intensity function $\lambda(x,y)$ for the process is given by
$$
\lambda(x,y)=\begin{cases}\lambda_{0} & (x,y)\in S \setminus C \\ \lambda_{1} & (x,y) \in C \end{cases},
$$
where $S$ and $C$ denote the square and the circle, respectively. The expected number of points in the pattern is then given by $\mu = \lambda_{0}(1-\pi / 10) + \pi\lambda_{1}/10$; the actual number of points in a given realization will be a stochastic variable distributed as $N \sim \mathrm{Poisson}(\mu)$.
Now, additivity tells us that if we have two Poisson distributed variables $X_{1}$ and $X_{2}$ with means $\mu_{1}$ and $\mu_{2}$, respectively, then their sum $X = X_{1}+X_{2}$ is also Poisson distributed with mean $\mu_{1} + \mu_{2}$. Moreover, if we condition on a particular total $x$, then we have
$$X_{1} \sim \mathrm{Binomial}\left(x,\frac{\mu_{1}}{\mu_{1}+\mu_{2}}\right),$$
and analogously for $X_{2}$. We will use this result to determine how many points to sample from the square and how many to sample from the circle.
To this end, let us decompose the intensity function into the sum of a function $\lambda_{S}(x,y)=\lambda_{0}$ defined on the entire square and a function
$$\lambda_{C}(x,y)=\begin{cases}\lambda_{1}-\lambda_{0} & (x,y) \in C \\ 0 & \text{otherwise}\end{cases}.$$ Therefore, if we want a total of $N$ points in our realization, we should draw a random integer distributed as
$$N_{S} \sim \mathrm{Binomial}\left(N,\frac{10}{10+\pi(\lambda_{1}/\lambda_{0} -1)}\right),$$
for the number of points to draw from the uniform distribution over the entire square, and then augment that with $N-N_{s}$ points from the uniform distribution over the circle. Note that the probability only depends on the ratio of the intensities between the "background" and the "feature", and not their absolute magnitudes. (A consequence of conditioning on a particular number of points in the pattern.)
So what to use for the ratio? Here things get a little fuzzy, as it is not entirely clear what you mean by points are "twice as likely" to be drawn from the circle as from the rest of the area. @A.G. and @Rojo have interpreted this to be a direct statement on the ratio of the intensities of the point process within these two areas, i.e., $\lambda_{1}/\lambda_{0} = 2$. On the other hand, @ubpdqn has interpreted this as meaning that 2/3 of the points fall within the circle, and your later post seems to favour this interpretation. These are not equivalent statements, as in the latter case, the binomial probability becomes $10/(30-3\pi)$.
So, let's code up both options (in the same function, no less):
generatePointSetConditionedOnNumberOfPoints[num_Integer,
opts : OptionsPattern[{"Interpretation" -> "TwoThirds"}]] :=
Module[{numBackground},
Which[OptionValue["Interpretation"] == "TwiceRate",
numBackground = RandomVariate[BinomialDistribution[num, 10/ (10 + Pi)]]
,
True,
numBackground = RandomVariate[BinomialDistribution[num, 10/(3 (10 - Pi))]]
];
Join[Table[generateUniformDistributedPointOnRectangle[{{0, 0}, {1, 1}}], {numBackground}],
Table[generateUniformDistributedPointOnCircle[{{1/2, 1/2}, Sqrt[1/10]}],
{num - numBackground}]]
]
In the above function, specifying the option "Interpretation"->"TwiceRate" gives the pattern corresponding to the interpretation used by @A.G. and @Rojo, while the (default option) "Interpretation"->"TwoThirds" corresponds to the choice by @ubpdqn.
Let's take a look at the difference on realizations consisting of 1000 points generated using each method:
pointSet1 =
generatePointSetConditionedOnNumberOfPoints[1000, "Interpretation" -> "TwiceRate"];
pointSet2 =
generatePointSetConditionedOnNumberOfPoints[1000, "Interpretation" -> "TwoThirds"];
GraphicsRow[
Map[ListPlot[#, Axes -> False, AspectRatio -> 1,
Epilog -> {EdgeForm[Black], FaceForm[None], Rectangle[],
Directive[GrayLevel[0.2], Dashed],
Circle[{0.5, 0.5}, Sqrt[1/10]]
},
PlotRange -> {{-0.025, 1.025}, {-0.025, 1.025}}] &,
{pointSet1, pointSet2}], Scaled[0.02], ImageSize -> 600]
The "twice the rate" interpretation is on the left, while the "two thirds of the points fall in the circle" interpretation is on the right. There is a fairly substantial difference between the two; in particular, the edge between the two regions is much harder to discern on the left than on the right. (Grey dashed circle added to help in identification.)
I apologize for the length; hopefully those brave souls who read the whole post found it useful and/or learned something along the way. And now ... off to bed.
