Distribution change of variables I'm working on distributions for a physics problem and I am quite stumped on how to proceed properly.
The problem is as follows: I start at the point $(0,0)$ and go a distance $\eta$ in x direction $(\eta, 0)$, from there I go in a random direction $\phi$ (uniformly distributed) a distance $x$ that is gaussian distributed. I need the distribution of $y$, with $y$ being the distance of the resulting point from $(0,0)$
To formalize that:
$f(x)=\frac{1}{b\sqrt{\pi}}\exp(-x^2/b^2)$
(is the distribution of the random x direction) but I also know that
$y = \sqrt{2 x \eta  \cos (\phi )+\eta
   ^2 + x^2}$
with $\phi$ being uniformly randomly distributed between $0, 2\pi$
Is it correct to say (with performing change of variables between x to y, and phi:
$g(y, \phi)\propto f(x)|\frac{dx}{dy}||\frac{dx}{d\phi}|$?
I'm pretty sure this is wrong (I suppose I need the Jacobian in some way), but I can't find good references for it

Whuber gave an answer to the question about the coordinate transformation. But in the final two sentences of his answer he noted

Finally, I suspect that this is just one step in the solution of a problem that, if it were disclosed, might be (much) simpler to solve some other way.  Analyzing the density of $(Y,\Phi)$ is not an appetizing prospect.

In your recent edit you disclosed that you are not after the joint distribution $f_{Y,\Phi}(y,\phi)$ but instead after the marginal distribution $f_Y(y)$.
In this answer we show one of those other more simple ways how to solve it without directly tackling the solution for the joint density distribution.

When I reverse engineer this, then I can model this as the distance $y$ between the origin and a random point on a circle with radius $\eta$ and midpoint $(0,x)$ where $x$ follows a normal distribution with standard deviation $b$.
Instead of computing the distribution density it might be easier to compute the cumulative distribution.
We can consider the values of $x$ and $\phi$ for which the point on the circle is still below a distance $y$ (ie. $f(x,\phi)<y$) and integrate over that area.
We can do this by considering for every point $x$ how much of the circle of radius $\eta$ centered at $0,x$ is intersecting with the circle of radius $y$ centered at $0,0$. Then integrate over all $x$ while multiplying with the fraction of the circle that intersects inside the circle of radius $y$.
$$F(Y<y) = 2 \int_{max(\eta-y,y-\eta)}^{y+\eta} \frac{1}{b\sqrt{\pi}} e^{-\frac{x^2}{b^2}} \overbrace{\frac{1}{\pi} cos^{-1}\left(\frac{x^2+\eta^2-y^2}{2x\eta}\right)}^{\text{fraction of circular segment intersecting}}dx + 2\int_0^{max(0,y-\eta)} \frac{1}{b\sqrt{\pi}} e^{-\frac{x^2}{b^2}} dx$$
To get the distribution density you differentiate the above. For this you can use Leibniz integral rule and you will get several terms because not only the integrand is dependent on $y$ but also the integration limits (but since the integrand is zero at the limits, or cancels, these terms will be zero as well).
If I do this then I get to
$$f(y) = 2 \int_{abs(y-\eta)}^{y+\eta} \frac{1}{b\sqrt{\pi}} e^{-\frac{x^2}{b^2}} \frac{1}{\pi} \frac{y}{x\eta}\frac{1}{\sqrt{1-\left(\frac{x^2+\eta^2-y^2}{2x\eta}\right)^2}}dx $$
or
$$f(y) = \frac{y}{b\pi^{3/2}} \int_{abs(y-\eta)}^{y+\eta}  e^{-\frac{x^2}{b^2}}  \frac{1}{\sqrt{x^2-(y-\eta)^2}\sqrt{(y+\eta)^2-x^2}}dx $$
Demonstration with computational simulation
Below is a histogram for a simulation of size $n=10^4$ with parameters $\eta= 1$ and $\sigma = b/\sqrt{2} = 1$. Superimposed is a computation of the density distribution based on a differentiation of the formula above.

### settings
set.seed(1)
n = 10^4
eta = 1
sig = 1

### simulate and plot histogram
phi = runif(n,0,2*pi)
x = rnorm(n,0,sig)
y = sqrt(eta^2+x^2+2*x*eta*cos(phi))

hist(y, freq=0, breaks = seq(0,5,0.1))

### integrand function
integrand = function(x,z) {
    ### exit for special cases to prevent NA values for acos
    if (x-eta>=z) {
      return(0)
    }
    if (x+eta<=z) {
      return(dnorm(x,0,sig))
    }
    return(dnorm(x,0,sig)*acos((x^2+eta^2-z^2)/(2*x*eta))/pi)
}
integrand = Vectorize(integrand)

### compute density by integrating integrand
pcomp = function(z) {
 d = 0.001
 xi = seq(abs(z-eta),z+eta,d)
 2*sum(integrand(xi,z))*d + 2*(pnorm(max(0,z-eta))-0.5)
}
pcomp = Vectorize(pcomp)

### compute cumulative distribution and differentiate for density
dd = 0.05
zs = seq(0,5,dd)
lines(zs[-1]-dd/2,diff(pcomp(zs))/dd,col=2)

