(a) General solution; speed constant; circular trajectory

Differentiate the first equation with respect to time and eliminate $v$ using the second equation:

$$\frac{\partial^2 u}{\partial t^2} - f \frac{\partial v}{\partial t} = 0,$$ (13)

$$\;\frac{\partial v}{\partial t} = - f u, \quad\Rightarrow\quad \frac{\partial^2 u}{\partial t^2} + f^2 u = 0.$$ (14)

Thus $u(t)$ satisfies a simple harmonic oscillator with frequency $\vert f\vert$:

$$u(t) = A\cos(ft) + B\sin(ft).$$ (15)

Using $\partial v/\partial t = - f u$ we integrate to get

$$v(t) = -A\sin(ft) + B\cos(ft) + C,$$ (16)

but a constant $C$ is excluded if we assume zero mean/component consistent with initial-value formulation; by direct substitution of initial conditions $u(0)=u_0$, $v(0)=v_0$ we find

$$u(t) = u_0\cos(ft) + v_0\sin(ft),$$ (17)

$$v(t) = -u_0\sin(ft) + v_0\cos(ft).$$ (18)

Notice that

$$u^2 + v^2 = u_0^2 + v_0^2 = U^2 = ,$$ (19)

so the speed $U(t)=\sqrt{u^2+v^2}$ is constant in time: inertial oscillations conserve kinetic energy when pressure and friction are absent.

The trajectory follows by integrating $dx/dt = u(t)$, $dy/dt = v(t)$. Using the above forms one obtains

$$x(t) - x(0) = \frac{u_0}{f}\sin(ft) - \frac{v_0}{f}(1-\cos(ft)),$$ (20)

$$y(t) - y(0) = \frac{u_0}{f}(1-\cos(ft)) + \frac{v_0}{f}\sin(ft).$$ (21)

These parametric equations describe a circle. The radius $R$ is

$$R = \frac{\sqrt{u_0^2 + v_0^2}}{\vert f\vert} = \frac{U}{\vert f\vert}.$$ (22)

Hence the trajectory is a circle of radius $U/f$ (using signed or absolute $f$ as appropriate).