See banked turn in aeronautics
In a motion along a circular path centered at $(0,0)$ (see figure below) with constant angular velocity $\omega$ the position of a point $P(x,y)$ (in Cartesian coordinates and $P(r,\varphi)$ in cylindrical coordinates) starting at $(r,0)$ at $t=0$ is given by the law of the uniform circular motion:
$x=r\cos(\omega t)$
$y=r\sin(\omega t),$
where $r$ is the radius of the circumference and $\omega=\dfrac{d\varphi}{dt}$. The magnitude of the tangential velocity is $v=\omega r$ and the radial velocity is zero. Thus $\omega=\dfrac{v}{r}$.
$x=r\cos\left(\frac{vt}{r}\right)$
$y=r\sin\left(\frac{vt}{r}\right)$
In your case $r=R$, $v=V$ (it is assumed that the aircraft mantains the speed $V$) and the centre of the circumference is $(X,Y)=(-R,0)$. If you make the change of coordinates $X=x-R$ and $Y=y$, you get the following $X,Y$ coordinates as a function of time $t$:
$X=R\left(\cos\left(\frac{V}{R}t\right)-1\right)$
$Y=R\sin\left(\frac{V}{R}t\right)$
In this $XY$-coordinate system the motion starts at $(X,Y)=(0,0)$ ($t=0$).
I assumed that the direction of the turn is to the left. If it is to the right, then one has a symmetric motion with respect to the $y-$axis:
$X=R\left(-\cos\left(\frac{V}{R}t\right)+1\right)$
$Y=R\sin\left(\frac{V}{R}t\right).$

The numerical result is obtained for $t=2$ and the other given data.
Added: For a different starting position $\left( X_{0},Y_{0}\right) $ at $t=0$ we have to incorporate these values in the motion equations. Integrating $% \omega =\dfrac{d\varphi }{dt}$, we obtain $\varphi =\omega t+C$, where $C$ is a constant. Then
$x=r\cos \left( \omega t+C\right) ,y=r\sin \left( \omega t+C\right) $
$x_{0}=r\cos \left( C\right) ,y_{0}=r\sin \left( C\right) $
and
$\dfrac{y_{0}}{x_{0}}=\dfrac{\sin \left( C\right) }{\cos \left( C\right) }% =\tan \left( C\right) ,C=\arctan \left( \dfrac{y_{0}}{x_{0}}\right) $
$x=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) ,y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) .$
Since
$X=x-r=r\cos \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}\right) \right) -r,Y=y=r\sin \left( \omega t+\arctan \left( \dfrac{y_{0}}{x_{0}}% \right) \right) $
we get
$X_{0}=x_{0}-r,Y_{0}=y_{0},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+r}$
$r=R,\omega =\dfrac{v}{r}=\dfrac{V}{R},\dfrac{y_{0}}{x_{0}}=\dfrac{Y_{0}}{X_{0}+R}.$
Finally, we obtain the equations of the motion of the aircraft in the $X,Y$-plane:
$X=x-R=R\left( \cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) -1\right) $
$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right),$
where $\left( X_{0},Y_{0}\right) $ is the position of the aircraft at $t=0$.
If the direction of the turn is to the right, then the motion is symmetric with respect to the $y-$axis:
$X=-x+R=R\left( -\cos \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R} \right) \right) +1\right) $
$Y=y=R\sin \left(\frac{V}{R}t+\arctan \left(\frac{Y_{0}}{X_{0}+R}\right) \right).$
Note: I changed the notation; in the question the position of the aircraft at $t=0$ is $\left( X,Y\right) $.
The velocity vector is $\overrightarrow{v}=v\overrightarrow{e}_{\varphi }$ and the acceleration vector $\overrightarrow{a}=-\dfrac{v^{2}}{r}\overrightarrow{e}_{r}$