Let $U$ be a simply connected bounded open set in $\mathbb{R}^2$. The area of $U$ is denoted by $A$.


(We do not assume any thing about its boundary).


Assume that $\gamma\_n$,s are smooth simple closed curves which lie in $U$.
The perimiter and area of $\gamma\_n$ are denoted by $l\_n$ and $A\_n$, respectively. We assume that $\gamma\_n$,s eventually leave compact subsets of $U$. That is for every compact subset $K\subset U$, there is a natural number $N$ such that $\gamma\_n$ has empty intersection with $K$, for every $n>N$.
Assume that $A\_n$ converges to $A$ and $l\_n$ converges to a real number $l$ and we have$ 4\pi A=l^2$.



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> Is $U$ necessarily the interior of a circle?
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An asymptotic version of the Isoperimetric inequality
I may have missed something but it should follow from [Bonnesen's inequality](https://en.wikipedia.org/wiki/Bonnesen%27s_inequality), which states that every domain $\Omega\subset\mathbb{R}^2$ satisfies :
$$\mathcal{L}(\partial\Omega)^2-4\pi\mathcal{A}(\Omega)\geq \pi^2(r\_\text{ex}(\Omega)-r\_\text{in}(\Omega))^2 $$ where $r\_\text{in}(\Omega)$ (resp. $r\_\text{ex}(\Omega)$) is the biggest( resp. smallest) possible radius of disk contained in $\Omega$ (resp. which contains $\bar\Omega$).


If one denotes by $\Omega\_n$ the domain bounded by your $\gamma\_n$. Then your hypothesis imply that each of the $\Omega\_n$ are sandwiched between two disks of closer and closer radius, which should be enough to conclude.


