Given $r>0$, find $k>0$ such that $\sqrt{(x-2)^2+(y-1)^2}Using the axioms, theorem, definitions of high school algebra concerning the real numbers, then prove the following:
Given $r>0$, find a $k>0$ such that:
$$\text{for all }x, y: \sqrt{(x-2)^2+(y-1)^2}<k\implies|xy-2|<r $$
I tried with several values given to $k$ and $r$ to find the relation between them.
Suppose then $r=1$ and we choose $k$ to be $\frac{1}{2}$
Now let us check if $\forall x,y$ such that $\sqrt{(x-2)^2+(y-1)^2}<\frac{1}{2}$ then $|xy-2|<1$
However we know that $|x-2|\leq\sqrt{(x-2)^2+(y-1)^2}$
Therefore if $\sqrt{(x-2)^2+(y-1)^2}<\frac{1}{2}$ then $|x-2|<\frac{1}{2}$..........(1)
For the same reason $|y-1|<\frac{1}{2}$...................(2)
But, (1) implies $\frac{3}{2}<x<\frac{5}{2}$.......................(3)
Also (2) implies $\frac{1}{2}<y<\frac{3}{2}$....................(4)
And the question now is, do ALL the values of x and y satisfying (3) and (4), satisfy $|xy-2|<1$
No because for $x=\frac{16}{10}$ and $y=\frac{6}{10}$ the relation $|xy-2|<1$ is not satisfied.
And the question is, which is the proper relation between $r$ and $k$ so that the above inequality is satisfied for all the values of $x$ and $y$

The center of the circle is at $(2,1)$, which is also on the hyperbola $xy = 2$. So it makes sense to write $x = 2 + a$ and $y = 1 + b$. Then your problem can be restated as: Given $r > 0$, find a $k > 0$ such that if $\sqrt{a^2 + b^2} < k$, then $|a + 2b + ab| < r$.
If $\sqrt{a^2 + b^2} < k$, then both $|a|$ and $|b|$ are less than $k$, which implies $|a + 2b + ab| < 3k  + k^2$. So your task is to find a $k$ such that $3k + k^2 < r$. 
I think you can take it from here...
