Distance between poisson points in two disjoint unit discs

Two disjoint unit discs $D_1$ and $D_2$. Inside them there are random poisson points with intensity $\lambda$. For a given real $r>0$, what is the probability that there exist a poisson point $x_1\in D_1$ and a poisson point $x_2\in D_2$ such that $\|x_1-x_2\|_2 \le r$? Couldn't find any existing results on this.

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How about 1D case, i.e., $D_1$ and $D_2$ are line segments? – user32828 Apr 19 '14 at 0:51

If $M$ is the $4$-dimensional Lebesgue measure of $\{(x,y) \in D_1 \times D_2: \|x - y\|_2 \le r\}$, then the number of such pairs of points is a Poisson random variable with parameter $M \lambda^2$, so the probability that there is at least one is $1 - \exp(-M \lambda^2)$.
EDIT: Consider the case where $D_1$ and $D_2$ are line segments, say $[0,1]$ and $[a,a+1]$ where $a > 1$. Of course we need $a-1 < r < a+1$ to make the question nontrivial. Let $X$ be the maximum of the Poisson points in $D_1$ ( $-\infty$ if there are none) and $Y$ the minimum of the Poisson points in $D_2$ ($+\infty$ if there are none). $X$ and $Y$ are independent, with densities $f_X(x) = \lambda e^{-\lambda(1-x)}$ and $f_Y(y) = \lambda e^{-\lambda(y-a)})$ on $[0,1]$ and $[a,a+1]$ respectively. Thus if $a-1 < r \le a$ $$P(Y - X \le r) = \lambda^2 \int_{a-r}^1 dx \int_a^{x+r} dy\; e^{-\lambda(1-x)} e^{-\lambda(y-a)} = {1 + \left( \lambda\,(a- r-1)-1 \right)\; {{ e}^{\lambda\, \left( a-r-1 \right) }}}$$ while if $a \le r < a + 1$ $$P(Y - X \le r) = (1-e^{-\lambda})^2 - \lambda^2 \int_0^{a+1-r} dx \int_{x+r}^{a+1} dy\; e^{-\lambda(1-x)} e^{-\lambda(y-a)} = 1 - 2 e^{-\lambda} + (1 - \lambda(a-r+1)) e^{\lambda(a-r+1)}$$
But how to compute $M$? – user32828 Apr 7 '13 at 11:58
Integrate over $x \in D_1$ the area of the intersection of $D_2$ with the disk of radius $r$ about $x$. I doubt that you'll end up with a closed-form expression. – Robert Israel Apr 8 '13 at 18:41