Consider a disk centered at the origin. Let $X$ and $Y$ be uniformly distributed on the disk. The conditional distribution of $Y$ given $X=x$ can be found as shown below.
$$f(x,y)=\frac{1}{\pi r^{2}},-\sqrt{r^{2}-x^{2}} \leq y\leq \sqrt{r^{2}-x^{2}}.$$
$$f_{Y|X}(y|x)=\frac{f(x,y)}{f_{x}(x)}.$$
$${f_{x}(x)}=\int_{a}^{b}f(x,y)dy=\int_{-\sqrt{r^{2}-x^{2}}}^{\sqrt{r^{2}-x^{2}}}\frac{1}{\pi r^{2}}dy = \frac{2\sqrt{r^{2}-x^{2}}}{\pi r^{2}}.$$
$$\frac{f(x,y)}{f_{x}(x)} =\frac{\frac{1}{\pi r^{2}}}{\frac{2\sqrt{r^{2}-x^{2}}}{\pi r^{2}}} = \frac{1}{2\sqrt{r^{2}-x^{2}}}.$$
I find it interesting that the area of the circle cancels out when calculating this conditional distribution. The conditional distribution only depends on the radius. This raises the question:
If we have a polygon with $n$ sides and an unknown area, can we use this previous calculation to derive its conditional distribution if $X$ and $Y$ are uniformly distributed on the space?
A general formula may look something like:
$$\frac{1}{Ubound-Lbound}.$$
Once you established a constant value $c$ of the variable $X$, you are considering a distribution of $Y$ over an intersection of your given space and a line $x=c$. If your space is
simply connecteda convex set (Thanks, @Rahul!) or at least normal with respect to the $X$ axis, the intersection is guaranteed to be a line segment, and the actual shape and area of the space does no longer matter – the initial distribution is uniform, so the resulting distribution is uniform. too, hence the density is just one over the length of the segment (and the length in the described case is $2\sqrt{r^2-x^2}$).