It is "intuitively" known and even taught in school that if a force with constant magnitude is applied perpendicular to a body's velocily, its trajectory will be circular. Let's explore this.
From Newton's 2-nd law: $\overrightarrow{F} = m \overrightarrow{a}$, so $ \Vert\overrightarrow{F}\Vert = const \implies \Vert\overrightarrow{a}\Vert = const$. Thus, if $\overrightarrow{v} = (v_x, v_y)$ then $(v_x')^2+(v_y')^2 = const \implies v_x'v_x''+v_y'v_y'' = 0 \ \ \ (1)$
Since $\overrightarrow{F} \perp \overrightarrow{v}$, $\ \ \ \ \overrightarrow{v} \cdot \overrightarrow{a} = (v_x, v_y) \cdot (v_x', v_y') = v_xv_x'+ v_yv_y' = 0 \ \ (2)$
From $(1)$ and $(2)$:
$ \left\{ \begin{array}{c} v_x'v_x''+v_y'v_y'' = 0 \\ v_xv_x'+ v_yv_y' = 0 \\ \end{array} \right. $
I don't really know how to solve this system. Maybe there's even some well-known method for that? I'm interested in rigorous solution with all the details why it's unique, not just "This function works."
Start with: $$v_xv^{'}_x + v_yv^{'}_y=0$$ This implies: $$v^{2}_x + v^{2}_y=R^2$$ for some constant $R$.
therfore it makes sense to parametrize as $$v_x = R\sin(\theta(t));v_y = R\cos(\theta(t))$$ and $$v^{'}_{x}=R\cos(\theta)\theta^{'};v^{'}_{y}=-R\sin(\theta)\theta^{'}$$
using $(v_x')^2+(v_y')^2 = const$ we obtain
$R^2(\theta^{'}(t))^2 = const$,
so $\theta(t)=\omega t +\theta_0$, and set $\theta_0=0$ w.l.g.
Now from $v_x(t)=R\sin(\omega t)$ we conclude that
$\{x,y\}=\{x_0,y_0\} + {R \over \omega} \{\cos(\omega t),-\sin(\omega t)\}$