Solution
= Solution
For $k=1$, the interval in part (b) contains one exactly when $a>1$. With $V=x^2+y^2=r^2$,
$$
\dot V=-2ar^2+4xy+2r^4
\leq2r^2(r^2-a+1),
$$
because $2xy\leq r^2$. Hence $\dot V<0$ throughout
$$
0<r^2<a-1.
$$
Every trajectory starting in the disc $x^2+y^2<a-1$ remains there and approaches the origin. Thus this disc is contained in the <basin of attraction> of the origin.