Uniform distribution over the unit circle Suppose that $U$ and $V$ are two independent uniform $(-1,1)$ random variables. 
Any hints on how I can show that their conditional distribution, given $U^2 +V^2<1$ is given by the uniform distribution over the unit circle, namely
$$f_{U,V|U^2+V^2<1} (u,v|w<1) =1/{\pi}, \quad u^2+v^2<1$$
Both an intuitive and a rigorous dervation would be helpful. Thank you.
 A: use the conditional probability formula $P(X \in A| X \in B) = \frac{P(X \in A \cap B)}{P(X \in B)}$ when $P(X \in B) > 0$.
Now take $X = (U, V), B = \{(u,v): u^2 + v^2 <1\}$. Let A be a measurable subset 0f $(-1, 1) \times (-1,1)$, since $P(X \in B) = \frac{m(B)}{4} = \frac{\pi}{4}$, we have $P(X \in A | X\in B) = \frac{\frac{m(A\cap B)}{4}}{\frac{\pi}{4}} = \frac{m(A \cap B)}{\pi}$. When A is a subset of B it's $\frac{m(A)}{\pi}$, which means uniform distribution ovr B.
More generally, if X is a uniform distribution over $B \subset R^n$, the conditional distribution knowing that X is in $A \subset B$ is the uniform distribution over A. This property helps to simulate some kinds of uniform distributions. For example, we want to simulate a uniform distribution in $x^2 + 2y^2 = 1$, then we could simulate a sequence of  uniform distributions in $[-1,1] \times [-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}]$ and take from the sequence the first point in $x^2 + 2y^2 = 1$ as the simulation result.
A: The original density of $(U,V)$ is constant on $\left(-1,1\right)^{2}$. Actually it equals $\frac{1}{4}$ since integration over that area must result in
$1$. The density of $(U,V)$ under condition $U^{2}+V^{2}<1$ is constant as well. There is only one choice then: $\pi^{-1}$. This because integration over that area must result in $1$ again.

$\int\int_{A}p_{U,V}\left(u,v\right)dudv=P\left\{ \left(U,V\right)\in A\right\} =1$ 

This for $A=\left(-1,1\right)^{2}$ in the original case and for $A'=\left\{ \left(u,v\right)\mid u^{2}+v^{2}<1\right\} $
in the conditional case. Both times with a constant density $p$.
A: We note that
$$
\begin{aligned}
\mathbb{P}(U< x,V< y|U^2+V^2<1)&=\frac{\mathbb{P}(U< x,V< y,U^2+V^2<1)}{\mathbb{P}(U^2+V^2<1)}\\
&=\frac{\mathbb{P}(U< x,V<y)}{\mathbb{P}(U^2+V^2<1)}\\
&=\frac{\mathbb{P}(U< x)\mathbb{P}(V<y)}{\mathbb{P}(U^2+V^2<1)}
\end{aligned}$$
Now let's apply total probability law to calculate
$$
\begin{aligned}
\mathbb{P}(U^2+V^2<1)&=\int_{-1}^1\mathbb{P}(u^2+V^2<1)f_U(u)du=\int_{-1}^1\mathbb{P}(V^2<1-u^2)\frac{1}{2}du\\
&=\frac{1}{2}\int_{-1}^1\mathbb{P}(|V|<\sqrt{1-u^2})du=\frac{1}{2}\int_{-1}^1\mathbb{P}(-\sqrt{1-u^2}<V<\sqrt{1-u^2})du\\
&=\frac{1}{2}\int_{-1}^1\left(\mathbb{P}(V<\sqrt{1-u^2})-\mathbb{P}(V<-\sqrt{1-u^2})\right)du\\
&=\frac{1}{2}\int_{-1}^1\left(\mathbb{P}(V<\sqrt{1-u^2})-\mathbb{P}(V<-\sqrt{1-u^2})\right)du\\
&=\frac{1}{2}\int_{-1}^1\left(\frac{\sqrt{1-u^2}+1}{2}-\frac{-\sqrt{1-u^2}+1}{2}\right)du\\
&=\frac{1}{2}\int_{-1}^1\sqrt{1-u^2}du\\
&=\frac{1}{2}\frac{\pi}{2}
\end{aligned}
$$
Therefore
$$
\mathbb{P}(U< x,V< y|U^2+V^2<1)=\frac{\frac{x+1}{2}\frac{y+1}{2}}{\frac{\pi}{4}}
$$
And finally the desired density function is 
$$
f_{U,V|U^2+V^2<1}(x,v|U^2+V^2<1)=\frac{\partial}{\partial x}\frac{\partial}{\partial y}\frac{(x+1)/2(y+1)/2}{\pi/4}=\frac{1}{\pi}
$$
