Ordered Points on a Line

Let $l$ be tangential to $\Gamma_2$ at $P$ ; so, triangle $APF$ is right angled at $P$ . Drop a perpendicular from $D$ onto $XY$ , intersecting $XY$ at the point $M$ . Triangles $AMD$ and $APF$ are similar, so, $\frac {DM}{PF} =\frac {AD}{AF}=\frac {3}{5}$ . Since $PF=15$ , so $DM =9$ .By Pythagorus, $XM^2= 15^2-9^2$ , so $XM =12$ . Thus, $XY= 2 \times XM =24$ .

[Latex edits - Calvin]

We drop a perpendicular $DZ$ upon $XY$ and we join $D,X ; F,W$ where $W$ is the point of contact of the tangent to the circle $\Gamma_2$ . Now, as F is the centre of the circle $\Gamma_2$ , we have $\angle FWA = 90^\circ$ . Therefore, $\triangle ADZ \sim \triangle AFW$ . So we have

$\dfrac{AD}{AF} = \dfrac{DZ}{FW}$ ,

or, $DZ = \dfrac{15 \times 3}{15 \times 5} \times 15$ = $\boxed{9}$

Now, in right angled triangle $DXZ$ , $DX^{2} = DZ^{2} + ZX^{2}$ . Putting the values of $DX=15 , DZ=9$ we have ,

$XZ=12= (\dfrac{1}{2} \times XY)$ . Therefore, $XY = \boxed{24}$

Let P and Q be feet of perpendiculars to the tangent respectively from D and F. Therefore for similar triangles
ADP and AFQ, $PD=AD*\dfrac {QF}{AF}=45*\dfrac{15}{75}=9$ . But this is a perpendicular from center D on
chord XY. Therefore $XY=2*\sqrt{15^2 - 9^2}=24.$
This is same as that of Anvesh Muttavarapu but a little different presentation.