Conditional joint density of $Z=Y/X$ So the question asks : Let X and Y be random variable with joint density: 
$f_{X,Y}(x,y) = c $ when $x^2 +y^2 ≤1 $ 
$f_{X,Y}(x,y) = 0 $ in other situations
(a) Find the value of the constant $c$.
(b) Let $Z=Y/X$. Compute the density of $Z$.
(c) Find the conditional density $f_{Y|X=x} (y)$. Identify the distribution of $Y |_{X=x}$.
So so far I got: 
a. $x^2+y^2<1$
so $x∈[-1,1]$ and $y∈[-(1-x^2)^{0.5},(1-x^2)^{0.5}]$
$1=\int_{-1}^{1} \int_{-(1-x^2)^{0.5}}^{(1-x^2)^{0.5}} c \, dy \,dx = cπ$
so $c = 1/π$
b. $F_Z(z) = P(Z ≤ z)= P(X/Y ≤ z) $
$=  \int_{-1}^1 \int_{ y / x ≤ z}^{xz} c \,dy \, dx $
$= 1/π \int_0^1 xz \, dx $
$= \left. z x^2 / 2 π \right|_0^1$
$= z / 2π $
so $f_{y/x}(z) = 1/2π $ for $-∞ ≤ z≤ ∞$
c. It is a uniform distribution on [$-(1-x^2)^{0.5} , (1-x^2)^{0.5}$]
It has a constant $c= 1/(b-a) = 2(1-x^2)^{-0.5}$
so $f_{y/x}(z)$ = $2(1-x^2)^{-0.5}$
I think the part a should be correct, but am I doing part b and c correctly? 
 A: Part (a) is correct.   Because $\pi$ is the area of the support, and total probability over this area must be $1$, a uniform distribution supported on the unit circle will have a probability density of: $1/\pi$. 

(b) Note that: $\mathsf P(Y/X\leq z) = \mathsf P(Y\leq Xz, X>1)+\mathsf P(Y\geq Xz, X<1)$
$$F_{Y/X}(z) = \int_{0}^{1}\int_{-\surd{(1-x^2)}}^{\min\{xz,\surd{(1-x^2)\}}} \pi^{-1}\operatorname d y\operatorname d x + \int_{-1}^{0}\int_{\max\{xz,-\surd{(1-x^2)}\}}^{\surd{(1-x^2)}} \pi^{-1}\operatorname d y\operatorname d x$$
But I'd use the ratio distribution formula:
$$\begin{align}f_{Y/X}(z) = \int \frac{\lvert x\rvert}{\pi} \mathbf 1_{x^2(1+z)^2<1}\operatorname d x\quad\mathbf 1_{z\in(-\infty;\infty)}
\\ = \int_{-1/\surd(1+z^2)}^{1/\surd(1+z^2)}\frac{\lvert x\rvert}{\pi}\operatorname d x\quad\mathbf 1_{z\in(-\infty;\infty)}
\\ = \dfrac{1}{\pi(1+z^2)}\quad\mathbf 1_{z\in(-\infty;\infty)}
\end{align}$$
And $F_{Y/X}(z) = \tfrac 1 2 + \frac{\arctan(z)}{\pi}$
-- In hind sight, we can see that conversion to polar coordinates might have been useful too, as $Y/X = \tan(\Theta)$ . --

(c)  Indeed: $f_{Y\mid X}(y\mid x) ~=~ \frac{1}{2\sqrt(1-x^2)}~\mathbf 1_{[x^2+y^2\leq 1]}$
You have argued that as the probability has uniform density over the disk, conditional probability over any curve segment within the disk has uniform density.   That is okay; but comes with a cautionary note.
Be very careful with this line of reasoning in the future as it is not always the case.   Of particular note is the Borel-Kolmogorov paradox, when dealing with great circles and uniform distribution over the surface of a sphere.
