Consider two fixed circles whose interiors are not disjoint with centers $A$ and $B$. Let a line $\ell$ be drawn perpendicular to line $AB$, which intersects both circles at $X$ and $Y$ respectively. The question is how to place $\ell$ to to maximize the length of $XY$. (Note that line $\ell$ intersects each circle twice; since we want to maximize $XY$ we choose $X$ and $Y$ so that they are on opposite sides of $AB$).
My conjecture is that we should place $\ell$ so that $AX$ and $BY$ are parallel but I am unsure of how to prove this.
Ideally I would have a geometric proof because the original statement is geometrical, and often, these optimization problems have a nice way of looking at them that makes it obvious where the maximal case occurs.
I would really just compute it analytically, and say the derivative is equal to $0$. I use $BY=b$, $AX=a$, $DB=x$. Then you get $YD=\sqrt{b^2-x^2}$. $DA=a-b-x$, so $DX=\sqrt{a^2-(a-b-x)^2}$. $$XY=\sqrt{b^2-x^2}+\sqrt{a^2-(a-b-x)^2}$$ $$\frac{d\, XY}{d\,x}=\frac{-x}{\sqrt{b^2-x^2}}+\frac{a-b-x}{\sqrt{a^2-(a-b-x)^2}}=0$$ You can rewrite this last equation as $$\frac{DB}{YD}=\frac{AD}{XD}$$ This implies that the triangles $BDY$ and $ADX$ are similar, and therefore $YB||AX$.