This problem can be solved using AM-GM inequality or the geometry method. Here, we present a trigonometry based solution. Let $r=x^2+y^2$. Then there exists $\theta$ so that $x=\sqrt{r}\cos\theta$ and $y=\sqrt{r}\sin\theta$. Setting these to the given relation leads to \begin{align*} 4r\cos^2\theta - 5r\sin\theta\cos\theta + 4r\sin^2\theta&=5\\ 4r(\cos^2\theta + \sin^2\theta) -5r\sin\theta\cos\theta &=5\\ 4r -\frac{5}{2}r\sin 2\theta &=5 \end{align*} $$\therefore\quad r=\frac{10}{8-5\sin 2\theta}$$ Now, it is clear that when $\sin 2\theta =1$, $r$ has a maximum value of $\boxed{\frac{10}{3}}$ and when $\sin 2\theta = -1$, $r$ reaches minimal at $\boxed{\frac{10}{13}}$.