In the Poincare disk, there exist points $A$ and $B$ on the same side of line $L$ such that no circle through them lies entirely on that side I am attempting this problem using Poincare disk.

I want to show for hyperbolic plane (using this Poincare disk), there exists 2 points $A$, $B$ lying on same side $S$ of line $L$ such that no circle through $A$, $B$ lies entirely within $S$.

My Attempt
I know  hyperbolic circles in the Poincare disk model are also Euclidean circles -  except that P-centers differ from Euclidean circle  when the center is not O.
It follows that for this reason, P-centers are closer (in a Euclidean sense) to the boundary circle than expected in the Euclidean case.
Accordingly, given any line L, I choose arbitrary points A,B such that A,B are close to the boundary and the line L. It remains to show that any circle through A,B intersects line L.

I am wondering if the reasoning is correct and how I could complete/improve it.
Thank you!
 A: Comment
Why is placement of circles these ways not possible?

A: A benefit of the Poincaré disc model is that it is embedded in the Euclidean (equivalently, complex) plane and we are free to use the latter's geometry (equivalently, algebra).
Any line in $\mathbb{D}^2$ can be mapped isometrically onto the segment of the imaginary-axis cutting the unit circle in $\mathbb{C}$, so WLOG take $L$ to be this "vertical" diameter of $\mathbb{D}^2$.
Choose any $z=x+iy\in\mathbb{D}^2$ such that the perpendicular-bisector between $z$ and the origin $O$ intersects the real-axis outside the disc, and $\lvert x\rvert \lt \lvert y\rvert$.
Figure $1$ shows the unique circle $C$ that goes through $z,O,\overline{z}$.

$\tag{Figure 1}$
Any circle through $z,\overline{z}$ whose centre lies to the right of $L$ in the above figure will clearly intersect $L$ and, by our choice of $y$, any circle whose centre lies on the interval $[x,0]$ will also intersect $L$, since the radius must be greater than $\lvert x \rvert$. Hence we need only consider circles centred on $(-\infty,x)$.
Figure $2$ shows what happens to the circle through $z, \overline{z}$ when you shift the centre of $C$ to the left or the right, i.e. any blue-shift circle no longer touches $L$ and any red-shift circle intersects $L$ in two points, which can be proved by an application of the $\mathbb{E}^2$-triangle inequality. But this means every circle through $z,\overline{z}$ of $\mathbb{E}^2$-radius $r\lt 1$, which includes all $\mathbb{D}^2$-circles through those points, must intersect $L$, and so we have found the necessary $A,B$ satisfying the criteria. $\qquad\square$

$\tag{Figure 2}$
