## Trigonometry 7th Edition

$(6,0), (0,-6)$
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 $(0,-6)$.