Six boys and six girls sit along a line alternately in $x$  ways; and along a circle again alternatively in $y$  ways, then

6 boys and 6 girls can be arranged alternatively in the following ways

$\boxed{\text{G}}\text{B}\text{G}\text{B}\text{G}\text{B}\text{G}\text{B}\text{G}\text{B}\text{G}\text{B}\boxed{\text{G}}$

Required number of ways $=2\times 6!\times 6!=x$

Circular permutation:

First arrange 6 boys in a circle in $\left(6-1\right)!=5!$  ways.  The gaps can be filled by 6 girls in ${}^6{P}_6=6!$  ways

$\therefore$  Number of ways $=5!\times 6!=y$

$\therefore \frac{x}{y}=\frac{2\times 6!6!}{5!6!}=12\Rightarrow \boxed{x=12y}$