Precalculus (6th Edition) Blitzer

See graph. $(0,-3)$ and $(2,-1)$
Step 1. See graph. Step 2. The two intersection points can be found as $(0,-3)$ and $(2,-1)$ Step 3. Plug-in the coordinates into the first equation; we have $0-(-3)=3$ and $2-(-1)=3$ showing that both points are on the line. Step 4. Plug-in the coordinates into the second equation: $(0-2)^2+(-3+3)^2=4$ and $(2-2)^2+(-1+3)^2=4$ showing that both points are on the circle.