Ok let's try. I think that there are a lot of way to do so and I will try one (maybe not the faster way, but should be clean enough). To understand the motion we need just two dimensions, so we work on a plane. We take a point and define its velocity $\vec{v}=(v_x,v_y)$ and say that it is subject to a constant (in module) acceleration that is orthogonal to $\vec{v}$, say $\vec{a}=(a_x,a_y)$. Let's fix an origin so that we can write
$$
v_x = v\cos\theta \qquad v_y=v\sin\theta
$$
where $\theta$ is an angle with respect to some origin point. Since $\vec{a}$ must be orthogonal, than it is in the form
$$
a_x = -a\sin\theta \qquad a_y=a\cos\theta
$$
so that $\vec{v}\cdot\vec{a}=0$ (you can also exchange the signs, is the same). To find the trajectory we write the equations of motion of the components which are
$$
\frac{d v_x}{dt} = a_x \qquad \frac{d v_y}{dt} = a_y
$$
Before starting doing derivatives, we can simplify the calculation proving that under our assumptions the module of the velocity $v$ is constant. Indeed
$$
\frac{dv^2}{dt}=\frac{d}{dt}\left(v_x^2+v_y^2\right)
=2\left(v_x\frac{dv_x}{dt}+v_y\frac{dv_y}{dt}\right)
=2\left(v_x a_x+v_y a_y\right) = 2\vec{v}\cdot\vec{a}=0
$$
so we proved that a perpendicular acceleration cannot vary the module of the velocity but only rotate it. Now we substitute the expression of the components in the equations of motion, obtaining
$$
\frac{d v_x}{dt}=-v\sin\theta\frac{d\theta}{dt}=-a\sin\theta \qquad
\frac{d v_y}{dt}=v\cos\theta\frac{d\theta}{dt}=a\cos\theta
$$
so
$$
\frac{d\theta}{dt}=\frac{a}{v}
$$
where the parameter on the right is a constant. Integrating
$$
\theta(t)=\frac{at}{v}+\phi
$$
where it is nice to define $\omega=a/v$ and $\phi=\theta(0)$ which is just the initial angle at $t=0$. Then, putting it into the definitions of the velocities we have
$$
v_x = v\cos(\omega t+\phi) \qquad v_y = v\sin(\omega t+\phi)
$$
Last step: integrate the positions. The coordinates are
$$
\frac{dx}{dt}=v_x \qquad \frac{dy}{dt}=v_y
$$
and it is easy to integrate them obtaining
$$
x=x_0+\frac{v}{\omega}\sin(\omega t+\phi) \qquad y=y_0-\frac{v}{\omega}\cos(\omega t+\phi)
$$
where $x_0$ and $y_0$ can be imposed by setting initial conditions. Notice that here you can recover the "standard" cosine-sine assignments just by adding a phase $\pi/2$ to $\phi$. Now, finally, we can recognize that this is a circular motion. Indeed, summing the square of the previous equations one finds that
$$
(x-x_0)^2+(y-y_0)^2=\left(\frac{v}{\omega}\right)^2
$$
which is the circle equation with center in $(x_0,y_0)$ and radius $R=|v/\omega|$.
