Answer
$\{(0,-6),(6,0)\}$
Work Step by Step
$x^{2}+y^{2}=36$ Equation $(1)$
$x-y = 6 $ Equation $(2)$
From Equation $(2)$
$-y = 6-x $
$y = x- 6 $
Substituting $y = x- 6 $ in Equation $(1)$
$x^{2}+y^{2}=36$
$x^{2}+(x-6)^{2}=36$
Using $(a-b)^{2} = a^{2} - 2ab+b^{2}$
$x^{2}+x^{2} -12x + 36 =36$
$x^{2}+x^{2} -12x + 36 -36 =0$
$2x^{2} -12x =0$
$2x(x-6) =0$
$x = 0$ or $x = 6$
Substitute $ x$ values in Equation $(2)$ to get $y$,
Let $x = 0$
$y = -6$
Let $x=6$
$y = 0$
$(0,-6)$ and $(6,0)$ satisfies both equations.
So, the solution set is $\{(0,-6),(6,0)\}$
