High school inequality find b in √[(x−1)^2+(y−2)^2]<b⟹|xy2−4|<a

[solakis1]

given a>0 find b>0 such that:
\(\displaystyle \sqrt{(x-1)^2+(y-2)^2}<b\Longrightarrow |xy^2-4|<a\)

 

[lfdahl]

My attempt:

Spoiler

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 $a = 4$:
$b = \sqrt{8}-2 \approx 0.8284$.