Assume that $X, Y$ are independent and have identical distribution on $(0, \infty)$. My idea of solution is:
If $X^2 + Y^2 = z$, then it means that $X$ can take all values from $0$ to $\sqrt{z}$. Then, from definition of conditional expectation, we would have: $$ \mathbb{E}(X \lvert X^2 + Y^2 = z) = \frac{1}{ \mathbb{P} (X^2 + Y^2 = z)}\int_{X \in (0, \sqrt{z})} X dP $$ where $dP$ is density of $X$.
If we have density of $X$, then computing $\mathbb{P}(X^2 + Y^2 = z)$ is doable.
Is this correct?