Answer
$(6,0), (0,-6)$
Work Step by Step
The line $x-y=6$ intersects the circle $x^2+y^2=36$ at $(6,0)$and$(0,-6)$ as can be seen from the figure.
The solution can also be obtained analytically by solving the system of equations formed by the two equations $x-y=6$ and $x^2+y^2=36$
$$\because x-y = 6 \,\,\, \therefore x = 6+y$$
Substituting for ($x=6+y$) in ($x^2+y^2=36$)
$$(6+y)^2+y^2 =36\\36+12y+y^2+y^2=36\\2y^2+12y=0\\2y(y+6)=0$$
$\therefore y =0$ and $y=-6$
$x|_{y=0} = 6+0=6$
$x|_{y=-6} = 6+(-6)=0$
$\therefore$ The coordinates of the points of intersection are $(6,0)$ and $(0,-6)$.
