Given these axioms:
- principle of inertia;
- $F = ma$;
- Newton's third law;
- Space is isotropic.
We have a body that moves along a straight line at speed $v_0$. Suddenly this body is tied to an ideally inextensible thread (see it as a centripetal force). How can I formally infer that motion is a circumference? What are the equations I have to impose whose solutions are all $(x, y)$ such that they are the place of the points of a circle? I'm trying, but what can not write are conditions where do somehow assumed that there must be a circle somewhere. I doubt that what I am trying to do cannot be deduced, but most likely I am grossly mistaken.
I don't believe the conditions you have given are sufficient to deduce that the motion will be circular. In the first place, you would have to specify that the string is straight and that its other end is anchored at a fixed point.
Even then, though, you can't deduce the path of the body will be a circle unless you specify that when the string gets attached to the body it is being held taut so that the tension at its outer end is $\ \frac{M v_0^2}{\ell}\ $, where $\ M\ $ is the mass of the body and $\ \ell\ $ the length of the string, and that the entire length of the string is already moving in a circle with angular velocity $\ \frac{v_0}{\ell}\ $. Under those conditions, the tension in the string at a distance $\ d\ $ from its fixed end will be $\ \frac{Mv_0 ^2}{\ell} + \frac{ v_0^4\int_d^\ell x\rho(x)dx}{\ell^2}\ $, where $\ \rho(x)\ $ is the density of the string at distance $\ x\ $ from its fixed end. After the string gets attached to the body, the tension along the entire length of the string will remain the same and the whole system of body plus string will continue to move in a circle with the same angular velocity.
Response to clarification of question by OP in comments below
In the more general case where the centripetal force is supplied by some unspecified mechanism, the path of the body cannot be deduced unless the magnitude of the force is completely specified as a function of time and the body's position.
If you use axes whose origin is the centre towards which the centripetal force is directed, and the body is at the point $\ (\ell, 0)\ $ at time $\ t=0\ $ when the force is first applied, and is moving parallel to $y$-axis with $\ \dot{y}(0) = v_0\ $, then the equations of motion of the body are \begin{eqnarray} M\ddot{x} &=& -\frac{Fx}{\sqrt{x^2+y^2}}\\ M\ddot{y} &=& -\frac{Fy}{\sqrt{x^2+y^2}}\ ,\\ \end{eqnarray} where $\ F\ $ is the magnitude of the applied force, and the initial conditions are: $$ x(0)=\ell,\ y(0)=0,\ \dot{x}(0)=0,\ \dot{y}(0) = v_0\ .$$ For the study of centripetal force problems in general, polar coordinates are much more convenient, but in this case, if $\ F\ $ remains fixed at the value $\ \frac{Mv_0^2}{\ell}\ $, you can easily verify by substitution that $\ x(t) = \ell\sin\left(\frac{v_0t}{\ell}\right)\ $ and $\ y(t) = \ell\cos\left(\frac{v_0t}{\ell}\right)\ $ satisfy the above equations of motion and initial conditions, so the body will continue moving with speed $\ v_0\ $ around the circumference of a circle of radius $\ \ell\ $. If the magnitude of the force is anything else, the path will not be a circle.