A point $(X, Y )$ is chosen randomly from the unit circle. Let $R = \sqrt{(X^2 +Y^2)}$. Find the joint density function of the random vector $(X, R)$ A point $(X, Y )$ is chosen randomly from the unit circle.
Let $R = \sqrt{(X^2 +Y^2)}$.
Find the joint density function of the random vector $(X, R)$
My try:
$R=\sqrt{X^{2}+Y^{2}} \implies 0\leq R\leq\sqrt{2} $
$F_{R}(r)=\begin{cases}
\frac{r}{\sqrt{2}} & \text{if }0\le r\le\sqrt{2},\\
0 & \text{if }r<0,\\
1 & \text{if }r>\sqrt{2}.
\end{cases}$
$F'_{R}=f_{R} \implies f_{R}(r)=\begin{cases}
\frac{1}{\sqrt{2}} & \text{if }0\le r\le\sqrt{2},\\
0 & \text{otherwise.}
\end{cases}$
First of all, is this correct ?
I suppose that i have to use this theorem: $f_{X|R}(x|r)=\frac{f(x,r)}{f_R(r)}$ , how can i find $f_{X|R}(x|r)?$
Thanks!

$R = \sqrt{X^2 + Y^2}$ where $X$ and $Y$ are uniformly distributed on a unit disk i.e. $x^2 + y^2 \leq 1$. So you should have $0 \leq r \leq 1$ and not $0 \leq r \leq \sqrt2$.
I will use the Jacobian method.
Use the fact that $ \displaystyle f_{XY}(x,y) = \frac{1}{\pi}~$ over unit circle.
As $r = \sqrt{x^2 + y^2}$, $y = \pm \sqrt{r^2 - x^2}$. First note that we must have $r \geq |x|$.
Now also note that we can multiply the density function by $2$ given the symmetry above or below $y = 0$. The jacobian itself is $ \displaystyle  \frac{r}{\sqrt{r^2 - x^2}}$.
$\displaystyle f(r, x) = 2 ~|J| ~f(x, y)$
i.e $ ~\displaystyle f(r, x) = \frac{2r}{\pi \sqrt{r^2-x^2}}, |x| \lt r \lt 1, -1 \lt x \lt 1$
