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 $y$

what you are looking for are the two hyperbolas that either touch at two place or touch and go through the end point of a diameter of the semicircle.  take the upper part of the semicircle $$(x-2)^2 + (y-1)^2 = k^2.$$ let the hyperbola $$xy = 2+r, \frac{dy}{dx} = -\frac y x$$ touch at $$x = 2 + \cos t, y = 1 + \sin t$$ the common tangent has the slope $-\frac{\cos t}{\sin t}.$  the constraints on $t$ are $$ (2+\cos t)(1+\sin t)= 2 + r, \frac{\cos t}{\sin t} =\frac{1+\sin t}{2+\cos t}$$ simplifying the last equation, we get $$ \cos t + 2 \sin t + \cos t \sin t= r,2\cos t-\sin t + \cos^2 t-\sin^2 t=0 \tag 1$$ 
here is what i will try to do. solve numerically $$2\cos t-\sin t + \cos^2 t-\sin^2 t=0$$ for $t.$ use the second equation $$r = \cos t + 2 \sin t + \cos t \sin t $$  to find $r.$
