What is the Lebesgue measure of a circle in $\mathbb{R}^2$
So my answer of zero. The way I do it is by putting the circle inside an annular ring and then shrinking the ring so that its measure (which is an upper bound to the circle) goes to zero. And therefore the measure of the circle is zero. Is this approach valid?
Yes, the approach is valid provided you already know that a closed disk of radius $r$ has measure $\pi r^2$. Since the circle of radius $r$ is contained in the set difference of disks of radii $r$ and $r-\epsilon$, it follows that its measure is at most $\pi(r^2-(r-\epsilon)^2)$, which can be made arbitrarily small.
If you haven't yet proved that a disk has measure $\pi r^2$, then use Fubini's theorem as GEdgar suggested.