We can use work-energy theorem to solve this problem.
1/2 m\|\vec v_f\|^2-1/2 m\|\vec v_i\|^2=\int_i^f \vec F\cdot d\vec x
In this equation, we notice that the force x dot an infinitesimal displacement == 0, since the force is always perpendicular to the speed, and therefore is perpendicular to the trajectory on which the particle moves. Using this fact we can obtain
1/2m(v_f^2-v_i^2)=0\\ \therefore \|v_f\|-\|v_i\|=0
As you can see, the magnitude of the speed remains stable, regardless of the equation of the force. Given this condition, it is easy to conclude that the motion is uniform circular motion because the length of the force is constant and centripetal (central).
It is true that you can arrive at the same conclusion by manipulating the equations of motion, but that might make things more complicated. I hope you find this answer helpful!
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