One vertex is $(x, y+r)$ or $(x, y-r)$. I'll assume the former (the latter case is similar, just swap $r$ and $-r$).
The vertices will have coordinates $(x+r\sin\theta,y+r\cos\theta)$, where $\theta$ is an integer multiple of $\frac{2\pi}{n}$. ($\frac{360}{n}$ if you prefer degrees to radians.)
As noted in Wikipedia's "Dodecahedron" entry, if $s$ is the length of an edge of a dodecahedron, and $r$ the radius of its circumsphere, then
$$r = s \frac{\sqrt{3}}{4}\left( 1 + \sqrt{5} \right)$$
So, if two points $A$ and $B$ are joined by and edge, and $O$ is the center of the dodecahedron, then $\triangle AOB$ is isosceles with legs $r$ and base $s$; applying the Law of Cosines to its vertex angle, we have ...
$$s^2 = r^2 + r^2 - 2 r\cdot r\cos\angle AOB = 2 r^2 \left( 1 - \cos\angle AOB \right) = 4 r^2 \sin^2\frac{1}{2}\angle AOB$$
so that
$$\sin\frac{1}{2}\angle AOB = \frac{s}{2r} = \frac{2s}{s\sqrt{3}\left(1+\sqrt{5}\right)} = \frac{\sqrt{3}\left(\sqrt{5}-1\right)}{6}$$
whence
$$\angle AOB = 2 \arcsin \frac{\sqrt{3}\left(\sqrt{5}-1\right)}{6} = 41.8103\dots^\circ$$
If $A$ and $C$ are non-adjacent vertices on a face, then $d := |AC|$ is a diagonal of a regular pentagon with side length $s$. Thus,
$$d = \frac{s}{2}\left( 1 + \sqrt{5} \right)$$
Just as above, we can compute
$$\sin\frac{1}{2}\angle AOC = \frac{d}{2r} = \frac{s\left(1+\sqrt{5}\right)}{s\sqrt{3}\left(1+\sqrt{5}\right)} = \frac{\sqrt{3}}{3}$$
whence
$$\angle AOC = 2 \arcsin \frac{\sqrt{3}}{3} = 70.5288\dots^\circ$$
(You may recognize this as the central angle between adjacent vertices of a cube. It's often helpful to realize that a dodecahedron's face diagonals form the edges of a family of cubes, as shown in the Wikipedia entry. Moreover, one can think of constructing a dodecahedron by taking a cube and pitching a pup-tent on each face, where a triangular tent face and a quadrilateral tent face form a regular pentagon.)
Best Answer
There is an inscribed cube in a regular dodecahedron, which will aid the calculation of the radius of the circumscribed sphere.
Assume the regular dodecahedron has edge length $1$. The edge of the cube is the diagonal of a unit regular pentagon:
$$\begin{align*} \text{Side of cube} &= \text{Diagonal of a pentagon face}\\ &= \frac{1+\sqrt 5}2\\ &= \varphi \end{align*}$$
The diameter of the circumscribed sphere is the space diagonal of that inscribed cube, which is just $\sqrt 3$ times the side length of the cube:
$$\begin{align*} \text{Diameter of sphere} &= \sqrt 3\cdot \text{Side of cube}\\ &= \sqrt 3\varphi\\ \text{Radius of sphere} &= \frac{\sqrt3}2\varphi \end{align*}$$
As in the diagram in the question, using cosine law with a triangle formed by the sphere centre and an edge of the dodecahedron:
$$\begin{align*} \cos\alpha &= \frac{r^2 + r^2 - 1^2}{2r^2}\\ &= \frac{2 - r^{-2}}{2}\\ &= \frac{2-\frac43\varphi^{-2}}{2}\\ &= \frac{2-\frac23(3-\sqrt5)}{2}\\ &= \frac{\sqrt5}3\\ \alpha &\approx 41.81^\circ \end{align*}$$