Motion under constant-magnitude force perpendicular to velocity I asked this on MathStackExchange, but now I think this problem is more appropriate for this site.
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 or just a simpler(but rigorous!) way to argue that the motion is circular? I'm interested in rigorous solution with all the details why it's unique, not just "This function works."
 A: In all the answers, everybody assumes the force field to behave like the magnetic force $\overrightarrow{F}=q \overrightarrow{V}  \times  \overrightarrow{B}$ with $\overrightarrow{B}$ constant.
You don't need to have $\overrightarrow{B}$ constant for the magnitude of the force to be constant. And as a result, the trajectory is not always a circle. If $\overrightarrow{B}(t)$ periodically flips from $\overrightarrow{B}$ to $-\overrightarrow{B}$, the magnitude of the force is conserved but the trajectory is a smooth curve made up of sectors of circles whose concavity alternate.

Let's try to solve the problem without assuming the force to be magnetic like.
The equations are:
$$\begin{cases} t=0~~\overrightarrow{V}(0)= \overrightarrow{V_{0}} \\ t>0~~ ||\overrightarrow{F}(t)||=F\\t \geq 0 ~~  \overrightarrow{F}. \overrightarrow{V}=0\\  m \frac{ d\overrightarrow{V} }{dt}= \overrightarrow{F} \end{cases} $$
This set of equations contains enough information to solve for both $\overrightarrow{F}$ and $\overrightarrow{V}$.
In geometry, the only isometric transformations are: the  translations, rotations, reflections and any combinations of them. For each of them, there is a force field whose norm is conserved.
Translation
This case corresponds to: $\overrightarrow{F}=\overrightarrow{constante}$
The integration of the motion equation is immediate:
$$\overrightarrow{V}(t)=\overrightarrow{V_{0}}+ \frac{t}{m}  \overrightarrow{F}$$
The condition:$\overrightarrow{F}. \overrightarrow{V}=0$ becomes:
$$\overrightarrow{F}. \overrightarrow{V_{0}}+\frac{t}{m}F^{2}=0$$
This equation must be true for any t, and it can only be so if $F=0$.
A case which is rather uninteresting.
Rotation
Rotations conserve the norm:
$$||\overrightarrow{F}(t)||^{2}=k~~ \Rightarrow \overrightarrow{F}. \frac{d\overrightarrow{F}}{dt} =0$$
$$~~ \Rightarrow \frac{d\overrightarrow{F}}{dt} =  \overrightarrow{ \omega} (t)   \times \overrightarrow{F}$$
Constant rotation: $\overrightarrow{ \omega} (t)=\omega\overrightarrow{z}$
The equation for $\overrightarrow{F}$ is easily solved:
$$  \overrightarrow{F}(t) = \overrightarrow{A}cos( \omega t)+  \big(\overrightarrow{z}  \times  \overrightarrow{A}\big)~sin( \omega t) $$
Since $\overrightarrow{A}$ is any constant vector, we can always orientate the space such that $\overrightarrow{A}$ is colinear to $\overrightarrow{e_{x}}$ .
$$ \overrightarrow{F}(t)= F~cos( \omega t)~ \overrightarrow{e_{x}}+  F~sin( \omega t) ~\overrightarrow{e_{y}}$$
The tip of $\overrightarrow{F}$ moves along a circle (circular polarization).
We can now plugin this expression for$ \overrightarrow{F}$ into Newton's second law to find $\overrightarrow{V}.$
$$m \frac{d \overrightarrow{V} }{dt}  = F~cos( \omega t)~ \overrightarrow{e_{x}}+  F~sin( \omega t) ~\overrightarrow{e_{y}}$$
The integration is immediate:
$$ \overrightarrow{V}(t)= \overrightarrow{V}(0)+ \frac{F }{ \omega.m } sin(\omega t)~\overrightarrow{e_{x}}-  \frac{F}{ \omega.m } ~ \big(cos( \omega t)-1\big)~\overrightarrow{e_{y}} $$
Let's check the orthogonality condition:$~~~~ t \geq 0 ~~  \overrightarrow{F}. \overrightarrow{V}=0$
$$At~~t > 0:~~~~\overrightarrow{F(t)}.\overrightarrow{V(t)}=0~~  \Leftrightarrow  ~~V_{y}(0)=- \frac{F}{m \omega } $$
So we find for the initial velocity:
$$  \overrightarrow{V_{0}}= -\frac{F}{m \omega } \overrightarrow{e_{y}} + V_{z}(0)\overrightarrow{e_{z}} $$
At last, we can integrate to find the position:
$$ \overrightarrow{OM}(t)=  \overrightarrow{OM}(0)+\overrightarrow{V_{0}}t-\frac{ F}{ \omega^{2}.m } \big( cos(\omega t)-1\big) \overrightarrow{e_{x}}  -  \frac{F}{ \omega^{2}.m } ~ \big(sin( \omega t)- \omega t\big) \overrightarrow{e_{y}} $$
That is:
$$\begin{cases}x(t)=x_{0}+\frac{F}{m~ \omega ^{2}}- \frac{F}{m~ \omega ^{2}} cos( \omega ~t)\\y(t)=y_{0}- \frac{F}{m~ \omega ^{2}} sin( \omega ~t)\\z(t)=z_{0}+V_{z}(0)~t\end{cases} $$
If: $$  \overrightarrow{V_{0}}= -\frac{F}{m \omega } \overrightarrow{e_{y}} + V_{z}(0)\overrightarrow{e_{z}} $$
Then the trajectory is an helix whose center's coordinates are:
$$\begin{cases}x_{c}=x_{0}+\frac{F}{m~ \omega ^{2}}\\y_{c}=y_{0}\\z_{c}=z_{0}\end{cases} $$
and radius:
$$R=\frac{F}{m~ \omega ^{2}}$$
The trajectory degenerates to a circle if $V_{z}(0)=0$. The noticeable point to observe is that there is only one possibility for the initial velocity, namely $V_{y}(0)=-\frac{F}{m \omega }$.
At last, the solution we just found can be recovered with a Laplace force type formula.
\begin{cases} \overrightarrow{F} =q \overrightarrow{V}  \times  \overrightarrow{B} \\ \overrightarrow{B}=B \overrightarrow{e_{z}}   \\  \omega = \frac{qB}{m} \end{cases}
A: with these equations you can solve this problem.
$$m\,\dot v_x=f_x+f_{cx}\quad (1)\\
m\,\dot v_y=f_y+f_{cy}\quad ~(2)\\
\dot v_x^2+\dot v_y^2=c\quad~~~~~~~~ (3)\\
v_x\,\dot v_x+v_y\,\dot v_y=0\quad (4)$$
where $~f_{cx}~,f_{cy}~$ are constraint forces
4 equations for 4 unknows $~f_{cx}~,f_{cy},~\dot v_x,~\dot v_y~$
you obtain
$$\dot v_x=v_y\,\sqrt{\frac{c}{v_x^2+v_y^2}}\\
\dot v_y=-v_x\,\sqrt{\frac{c}{v_x^2+v_y^2}}$$
this is the numeric solution

analytical solution
assume that
$$v_x^2+v_y^2=\text{constant}=d$$
you obtain
$$\dot v_x=v_y\,\sqrt{\frac{c}{d}}=a\,v_y\\
\dot v_y=-v_x\,\sqrt{\frac{c}{d}}=-a\,v_x$$
the solutions
$$x(t)={\frac {v_{{{\it y0}}}+x_{{0}}a}{a}}+{\frac {v_{{{\it x0}}}\sin
 \left( at \right) }{a}}-{\frac {v_{{{\it y0}}}\cos \left( at \right) 
}{a}}
\\
y(t)={\frac {v_{{{\it x0}}}\cos \left( at \right) }{a}}+{\frac {v_{{{\it y0
}}}\sin \left( at \right) }{a}}+{\frac {-v_{{{\it x0}}}+y_{{0}}a}{a}}
$$
this is a circle equation because
$$\left( x \left( t \right) -{\frac {v_{{{\it y0}}}+x_{{0}}a}{a}}
 \right) ^{2}+ \left( y \left( t \right) -{\frac {-v_{{{\it x0}}}+y_{{0
}}a}{a}} \right) ^{2}
={\frac {{v_{{{\it x0}}}}^{2}+{v_{{{\it y0}}}}^{2}}{{a}^{2}}}=\rho^2$$
your equations
$
\left\{ 
\begin{array}{c}
v_x'v_x''+v_y'v_y'' = 0 \\ 
v_xv_x'+ v_yv_y' = 0 \\ 
\end{array}
\right. 
$
you can check that the solution $~x(t),~y(t)~$ fulfilled also your equations. where
$$'=\frac{d}{dt}$$
A: We choose a frame of reference in which the $ z $ axis is perpendicular to the plane containing the force and the velocity at a given instant ie in the direction $\mathbf{ F} \wedge \mathbf{v} $. Since there is no force in this direction, the velocity  $ v_{z} $ will remain zero.
It is shown below (see Lemma 1) that since $ \mathbf{ F(t)} $ is perpendicular to $ \mathbf{v} $
\begin{equation}
 \mathbf{F}(t) = K (v_{y}(t), - v_{x}(t))   
\end{equation}
where $ K $ is a constant.
Hence we have
\begin{equation}
  \mathbf{F}(t)= m \mathbf{\dot{v}} = m (\dot{v}_{x}(t),  \dot{v}_{y}(t))  = K (v_{y}(t), - v_{x}(t)) 
 \end{equation}
where $m$ is the mass of the particle. Thus
\begin{eqnarray}
 m \dot{v}_{x}(t) &  = & K v_{y}(t)\\
 m \dot{v}_{y}(t) & = & - K v_{x}(t)
\end{eqnarray}
A simple integration gives
\begin{eqnarray}
 v_{x}(t) &  = & (K/m) y(t) - a \\
 v_{y}(t) & = & - (K/m) x(t) +b
\end{eqnarray}
where $ a $ and $  b $ are constants.
But we know that $ \mathbf{v}.\mathbf{v} = constant $ and hence we have
\begin{equation}\label{key}
 ((K/m) y(t) - a)^{2} + (- (K/m) x(t) +b)^{2}  = constant
\end{equation}
which can be trivially rearranged into
\begin{equation}
 (y(t) - A)^{2} + (x(t) - B)^{2}  = R^2
\end{equation}
where A, B and R are constants. This is a circle of radius R centred on (A,B) and is the desired result.
Lemma 1
If $ \mathbf{F}.\mathbf{v} = 0 $,  it follows that $ F_{x} v_{x} + F_{y} v_{y} = 0 $ and hence by simple re-arrangement that $ F_{x} = - F_{y} v_{y} / v_{x} $. Thus $ \mathbf{F} = -\frac{F_{y}}{v_{x}}( v_{y}, - v_{x}) $
Since $ \mathbf{F}.\mathbf{F} = c_{1} = $constant,
$\mathbf{F}.\mathbf{F} = (-\frac{F_{y}}{v_{x}})^2 (( v_{y}, - v_{x}).( v_{y}, - v_{x})) = c_1$
Thus
$\mathbf{F}.\mathbf{F}=(-\frac{F_{y}}{v_{x}})^2 ( v_{y}^2+v_{x}^2)= (-\frac{F_{y}}{v_{x}})^2   c_2$
where $ \mathbf{v}.\mathbf{v} = (v_{y}^2+v_{x}^2)=c_{2}$, a constant.
ie
$c_1=(-\frac{F_{y}}{v_{x}})^2 c_2$
Since $c_1$ and $c_2$ are constants, it follows that $ - \frac{F_{y}}{v_{x}}$ must be also be a constant which we call  $ K $. Hence we have
\begin{equation}
 \mathbf{F} =  K (v_{y}(t), - v_{x}(t)) 
\end{equation}
as is used above.
A: Since the force is perpendicular to velocity, it does not work, so speed $v = |\vec v|$ is conserved:
$$
\vec v\cdot \vec a = 0\\
\frac{d}{dt} \frac{1}{2}v^2 = 0
$$
This means you only need to solve for the angle of the velocity $\phi$. You have using polar coordinates:
$$
mv\dot \phi = F
$$
which you can easily solve for
$$
\phi = \phi_0+\frac{Ft}{mv}
$$
ie the velocity vector executes a uniform circular motion at angular frequency $\omega = \frac{F}{mv}$ (which depends on the initial condition).
You'll need to integrate this to obtain the position, which is straight forward, especially if you use complex numbers. You'll find the combination of a uniform circular motion at same frequency, accompanied by a uniform rectilinear drift.
Hope this helps
A: Let's try for the simpler but rigorous argument.
There is a forward velocity v, which can change with time.
At any given time, there is a force with a constant magnitude, which is perpendicular to the velocity at that time. The force is perpendicular to the velocity, so it does not change the speed. It's assumed but not stated that nothing else changes the speed either. It's unstated but assumed that the mass the force acts on is constant, so the magnitude of the acceleration is also constant.
Since the acceleration relative to the current velocity is constant, in a given time the amount of deviation from the current direction of velocity is also constant. We don't have to calculate how much that is, only that it's constant. This is what it means to have a constant angular velocity.
Constant speed, constant angular velocity. What shape can that give us but a circle? Show me a spot where the tangent to the path is different from the tangent for another point on the path, and I'll show you a point where the speed or the angular velocity is not the same.
This may not be rigorous, but I hope it's convincing.
