Let  $L_1$ be the length of the common chord of the curves $x^2+y^2=9$ and  $y^2=8x,and{L}_2$  be the length of the latusrectum of $y^2=8x,$ then

$x^2+y^2=9and{y}^2=8x$

$\begin{array}{l}\Rightarrow x^2+8x-9=0\\ \Rightarrow \left(x+9\right)\left(x-1\right)=0\\ x=-9,x=1\end{array}$

$\begin{array}{l}x=-9\Rightarrow rejected\\ x=1\Rightarrow y^2=8\\ \Rightarrow y=\pm 2\sqrt{2}\end{array}$

Length of Common chord $=2y=4\sqrt{2}$

$y^2=8x$$\Rightarrow$Length of latus rectum = 4a = 8

$\begin{array}{l}\Rightarrow L_2=8\\ \therefore L_1<L_2\end{array}$