The $x$ and $y$ values are confined to the area of an open disc of radius $b$ with center at $(1,2)$.
The largest allowed $y$-variation ($y_{max}-y_{min} = 2b$) happens for $x=1$.
The largest allowed $x$-variation happens for $y=2$.
Hence, we have the relations:
$x=1$:
$|y-2| < b\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$(1).
and $|y^2-4| < a\;\;\;\;\;\;\;$(2).$y=2$:
$|x-1| < b\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$(3).
and $|4x-4| < a$ or $|x-1| < \frac{a}{4}\;\;$(4).(2). Implies: $\sqrt{4-a}<y <\sqrt{4+a}$. Note, that $a \le 4$. (There is a symmetric case for $y < 0$, which is omitted).
(1). Implies: $2-b < y < 2+b$.
Now, $a >0$ is given, therefore:
$2+b \le \sqrt{4+a}$ and $2-b \ge \sqrt{4-a} $
$\Rightarrow b \le 2- \sqrt{4-a}$ and $b \leq \sqrt{4+a}-2$.
We need to choose the smaller of the two, which is the latter: $\sqrt{4+a}-2$. This can be seen e.g. from their Taylor expansion:
\[2- \sqrt{4-a} \approx \frac{a}{4}+\frac{a^2}{64}+O(a^3) \\\\ \sqrt{4+a}-2 \approx \frac{a}{4}-\frac{a^2}{64}+O(a^3)\]
From (3) and (4) we immediately get: $b \le \frac{a}{4} $.
The difference between the two possible boundaries - according to the Taylor expansion - is:
\[\sqrt{4+a}-2 -\frac{a}{4} \approx -\frac{a^2}{64}+O(a^3) < 0\]Thus, $b =\sqrt{4+a}-2$ is the only possible choice. The largest possible $b$-value is obtained when $a = 4$:
$b = \sqrt{8}-2 \approx 0.8284$.