Solution of simultaneous ODE : $x'=(x^2+y^2)y$ and $y'=-(x^2+y^2)x$
You can assemble the two equations into a complex-scalar equation as $$ \dot z=-i|z|^2z,~~~z=x+iy. $$ As observed, the radius has then the equation $$ r\dot r=\frac12\frac{d}{dt}|z|^2=Re(\bar z\dot z)=Re(-i)|z|^4=0, $$ giving $r=c$ constant. Then the original equation has a simple linear form $$ \dot z=\lambda z,~~~\lambda =-ic^2\\ \implies z(t)=z_0e^{-ic^2t},~~~|z_0|=c\text{ or }z_0=ce^{i\phi}\\ \implies x(t)+iy(t)=c\,\Bigl(\cos(c^2t-\phi)-i\sin(c^2t-\phi)\Bigr) $$