Secant chord on intersecting circles construction "The secant circles $Γ_1(O_1,R_1)$ and $Γ_2(O_2,R_2)$ intersect at points $A$ and $B$. Given a line of lenght $l$, explain how to construct a straight line passing through $A$ intersecting $Γ_1$ and $Γ_2$ at points $X$ and $Y$ respectively , such that $\overline{XY}=l$. Under what conditions is there a solution?"
I tried to solve it but I couldn't and the solution on the textbook doesn't seem right.What are the steps and the conditions so that there is a solution?
The solution from the textbook:"Assuming that the problem has been solved, let $M$ and $N$ be the midpoints of line segments $\overline{AX}$ and $\overline{AY}$, respectively, so that $\overline{XY}=\frac{l}{2}$. Construct triangle
$O_1O_2P$, right at $P$ and such that $\overleftrightarrow{O_1P}‖\overleftrightarrow{MN}$. Show that there will be a solution if and only if $O_1O_2\geq\frac{l}{2}$." I measured the possible lengths of the chords and for $\overline{O_1O_2}=4cm$, an $XY$ measuring more than $8cm$ worked as a possible secant as well as some smaller values as indicated by the solution/hint.
 A: Have your tried drawing this? You should have something like the following situation as a mental model:

There is one mistake in the textbook answer: you want $\overline{XY}=l$ but $\overline{MN}=\frac l2$.
You can see from the image that $M$ and $N$ are the orthogonal projection of $O_1$ and $O_2$ onto the line $XY$. This is due to e.g. $\triangle O_1AX$ being isosceles. So you want to project the two points in such a direction that their projected distance is $\frac l2$, which is what $\triangle O_1O_2P$ achieves. This is only possible if their unprojected distance is at least that big. Once you have that triangle, take the parallel to $O_1P$ through $A$ to find $X$ and $Y$.
If you claim that for $\overline{O_1O_2}=4\text{cm}$ you got some $\overline{XY}>8\text{cm}$ in contradiction to the textbook solution, it would be really interesting to see a picture for this and work out where you went wrong. In particular, is that secant still passing through $A$? From the considerations above, you get maximum distance if $\overleftrightarrow{O_1O_2}\parallel\overleftrightarrow{XY}$ and in that case you get $\overline{XY}=2\,\overline{O_1O_2}$. I suspect there must have been some oversight, misunderstanding or inaccuracy in your drawing for this.
For the sake of completeness: the other extremal situation is $\overleftrightarrow{O_1O_2}\perp\overleftrightarrow{XY}$ which leads to $X$ and $Y$ both coindicing with $B$, giving $l=0$.
Update: Providing details on how to construct $\triangle O_1O_2P$. You want $P$ to be on a circle with diameter $O_1O_2$. According to Thales's theorem this gives you a right angle at $P$. You also want $P$ on a circle of radius $\frac l2$ around $O_1$ to get $\overline{O_1P}=\overline{MN}=\frac l2$. Intersect these two circles and you have $P$. Then you draw a parallel to $\overleftrightarrow{O_1P}$ through $A$ and you have the line $\overleftrightarrow{XY}$. Intersect it with the circles and the intersections that are not $A$ will be $X$ and $Y$ respectively.
There is a lot of arbitrary choices here. There are two points of intersection between the two initial circles. They lead to two different solutions for $X$ and $Y$. You could also swap the roles of $O_1$ and $O_2$, i.e. draw a circle around $O_2$ with radius $\frac l2$. That will lead to two different positions for point $P$, but the directions of the lines connecting these points to $O_2$ will match the two earlier solutions we had for $O_1$. So in total there are two possible solutions for the line $XY$.
Here is a figure with the first two circles and all these alternate solutions drawn in:

