The difference of squares of two numbers is $180$. The square of the smaller number is $8$ times the larger number. Find the two numbers.

(A) Let the larger number be $x$ and the smaller number be $y$.
According to the given conditions:
$x^2 - y^2 = 180$ --- $(1)$
$y^2 = 8x$ --- $(2)$
Substituting equation $(2)$ into equation $(1)$:
$x^2 - 8x = 180$
$x^2 - 8x - 180 = 0$
Factoring the quadratic equation:
$x^2 - 18x + 10x - 180 = 0$
$x(x - 18) + 10(x - 18) = 0$
$(x - 18)(x + 10) = 0$
So,$x = 18$ or $x = -10$.
Since $y^2 = 8x$,if $x = -10$,then $y^2 = -80$,which is impossible for real numbers. Thus,$x = 18$.
Now,find $y$:
$y^2 = 8(18) = 144$
$y = \pm \sqrt{144} = \pm 12$.
Therefore,the two numbers are $(18, 12)$ or $(18, -12)$.