For simplicity don't put the side on the $x$ axis; put the center at the origin $(0,0)$.
The polygon is enscribed in a circle. For simplicity we can assume the circle has radius $1$ and each of the vertices will be $(\cos \theta, \sin \theta)$ for some angle $\theta$.
Assuming the polygon is an $n$-gon, The $n$ radii connected to the center create $n$ central angles of $\frac {2\pi}{n}$ radians or $\frac {360^{\circ}}{n}$ degrees.
If we assume vertex $v_0$ is at $(1,0)$ then the remaining vertices $v_k$ will be $(\cos k\frac {2\pi}{n}, \sin k\frac {2\pi}{n})$ for $k = 0,.... , n-1$.
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If we scale the $n$-gon by a scale of $R$ then the vertices will be $(R\cos k\frac {2\pi}{n}, R\sin k\frac {2\pi}{n})$.
If we rotate the $n$-gon by an angle of $A$ then the vertices will be $(R\cos (k\frac {2\pi}{n}+A), R\sin (k\frac {2\pi}{n}+A))$
If we move the center of the $n$-gon to point $(X,Y)$ then the vertices will be $(X + R\cos (k\frac {2\pi}{n}+A), Y+R\sin (k\frac {2\pi}{n}+A))$
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So you instead you are given that one vertex is $(x_1,y_1)$ and the (counter-clockwise) adjacent vertex is $(x_2,y_2)$
The you need to solve $X,Y, R, A$ from:
$(x_1,y_1) = (X + R\cos (0\frac {2\pi}{n}+A), Y+R\sin (0\frac {2\pi}{n}+A))= (X + R\cos A, Y + R\sin A)$
and $(x_2, y_2) = (X + R\cos (\frac {2\pi}{n}+A), Y+R\sin (\frac {2\pi}{n}+A))$
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Some things that may help: The side of a unit $n$-gon is $2\sin \frac {\pi}{n}$ so $R = \frac {\sqrt{(x_2-x_1)^2 - (y_2-y_1)^2}}{2\sin \frac {\pi}{n}}$.
The angle of the side of a unit $n-$gon from $(1,0)$ to $(\cos \frac {2\pi}n, \sin \frac {2\pi}n)$ is $\pi - \frac {\pi - \frac {2\pi}n}2=\frac \pi 2 + \frac \pi n$. So $A=\arctan \frac{y_2-y_1}{x_2-x_1} - \frac \pi 2 -\frac \pi n$.