Answer
$(2+\sqrt 7,2+\sqrt 7)$ and $(2-\sqrt 7,2-\sqrt 7)$
Work Step by Step
Step 1. Points that are 4 units from $(1,3)$ are on a circle: $(x-1)^2+(y-3)^2=16$
Step 2. Let $y=x$ in the equation, we have $(x-1)^2+(x-3)^2=16$ or $x^2-4x-3=0$ which gives $x=\frac{4\pm\sqrt {16+4(3)}}{2}=\frac{4\pm\sqrt {28}}{2}=2\pm\sqrt 7$
Step 3. The points are $(2+\sqrt 7,2+\sqrt 7)$ and $(2-\sqrt 7,2-\sqrt 7)$
