Variation of Radon transform for probability measures on $\mathbb C$ Let $\mu$ be a probability measure on $\mathbb C$. For $z \in \mathbb C$, let $$f^z \colon \mathbb C \to \mathbb R_{\geq 0}$$ be the function $f^z(\lambda) = |\lambda - z|$. Consider now the family $(\mu_z)_z$ of probability measures on $\mathbb R_{\geq 0}$ with $\mu_z := f^z_*(\mu)$, i.e., the push-forward of $\mu$ with respect to $f^z$.

Question: Does the family $(\mu_z)_z$ determine $\mu$?

I am convinced that the answer to the previous question must be positive, but I do not see a direct argument. Maybe it is enough to know the measure of disks of radius $1$ around each $z \in \mathbb C$. I am more interested in the second question: 

Question: What consistency condition must the family $(\mu_z)_z$ satisfy to ensure that a suitable measure $\mu$ exists?

Last question:

Question: Given $(\mu_z)_z$ coming from this construction, is there some inversion formula that describes $\mu$ in terms of $(\mu_z)_z$.

 A: Let me turn my comment into an answer.
I claim that
for fixed $\mu\in\text{Prob}(\mathbb{C})$ and $f\in C_c(\mathbb{C})$,
$$ \int f(z)d\mu(z)= \lim_{r\to 0} \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w), $$
where $\lambda$ is the Lebesgue measure on $\mathbb{C}$.
To see this,
fix $\epsilon>0$, let $\delta$ be the one given by the uniform continuity of $f$, and observe that for $r\leq \delta$,
$$ \left|\int f(z)d\mu(z)- \frac{1}{\pi r^2} \int\mu_w([0,r))f(w) d\lambda(w)\right|=$$
$$ \left|\int d\mu(z)\int_{|w-z|\leq r} d\lambda(w)\frac{f(z)}{\pi r^2} - \int d\lambda(w)\int_{|z-w|\leq r} d\mu(z)\frac{f(w)}{\pi r^2} \right| =$$
$$\left| \int d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{f(z)-f(w)}{\pi r^2}\right|\leq$$
$$ \int d\mu(z)\int_{|w-z|\leq r}d\lambda(w) \frac{|f(z)-f(w)|}{\pi r^2} \leq\epsilon.$$

This gives the required inversion formula, hence answers Q3, as well as Q1.

As for Q2, I am not sure what do you expect. One way to answer would be using the inversion formula given above, that is given a collection $(\mu'_z)_z$ you can define $\mu$ using the above formula and then define the collection $(\mu_z)_z$ as you did. The initial collection is "consistent" iff you got back where you've started.
A: The answer to Q1 is yes, as you suspected. This follows from a sufficiently general version of Lebesgue's differentiation theorem (or Vitali's covering theorem). See, for example, Mattila, Geometry of sets and measures in Euclidean spaces, Theorems 2.8, 2.12.
If $\mu,\nu$ are two Borel measures on $\mathbb C$, then
$$
f(z) = \lim_{r\to 0+} \frac{\mu(D_r(z))}{\nu(D_r(z))}
$$
exists for $\nu$-almost every $z\in\mathbb C$ and computes the Radon-Nikodym derivative of the absolutely continuous (with respect to $\nu$) part of $\mu$. In particular, $\mu(B)\le \int_B f(z)\, d\nu(z)$ for every Borel set $B\subseteq\mathbb C$. Since $f=1$ if $\mu,\nu$ agree on disks, this gives that $\mu(B)\le \nu(B)$, so $\mu=\nu$ by symmetry.
