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 $\overrightarrow{a}=-\dfrac{v^{2}}{r}\overrightarrow{e}_{r}$
