finding joint probability density function of a point inside a circle I'm pretty stumped with this problem that I am working on. Any help on how to go about it?
A point is chosen at random from the interior of a circle whose equation is x^2 + y^2 ≤ 4. Let the random variables X and Y denote the x- and y-coordinates of the sampled point. Find fx,y(x, y).
logically, the answer would seem to be 1/4pi given that it should be 1/(area of circle). I'm not sure if:
a) that is correct
b) what the proper way is to go about solving it using traditional methods
thank you! 
 A: Let $D=\{(x,y) \in \mathbb{R}^2\mid x^2 + y^2 \le 4\}$.

Assuming the distribution is uniform on $D$, the density is constant on $D$, equal to $c\;$say, for some positive real number $c$, and zero outside of $D$.

Hence, the necessary condition on $c\;$is
$$\iint_{D} c\,dA= 1$$
so we get
$$1 = \iint_{D}c\,dA = c \iint_{D} 1\,dA = c \times (\text{the area of $D$}) = c(4\pi)$$
which implies $c = {\large{\frac{1}{4\pi}}}$.
A: Let $f(x)$ be the p.d.f. such that $f(x)dx$ gives the probability that a point inside the circle is chosen whose X-coordinate lies in the neighborhood of $x$. Obviously $f(x)$ is non-zero only for $-2\leq x\leq 2$. Since the probability is uniformly distributed over the area of the circle $f(x)dx$ must equal the area of the strip of width $dx$ at $x$, which is equal to $2\sqrt{4-x^2}dx/(4\pi)$. Hence $f(x)=2\sqrt{4-x^2}/(4\pi)$. 
Now for a given value of $x$, we must have $-\sqrt{4-x^2}\leq y\leq +\sqrt{4-x^2}$. Again since probability is uniformly distributed over this line segment, the conditional p.d.f. $f(y|x)=1/(2\sqrt{4-x^2})$.
Now $f(x,y)=f(x)f(y|x)=1/(4\pi)$.
