For part b, I'm getting a little stuck. So I'm trying to show that if $(x,y,r)R(a,b,s)$ and $(a,b,s)R(x,y,r)$, then $(x,y,r)=(a,b,s)$
And so $(x,y,r)R(a,b,s)$ implies $\sqrt{(x-a)^2+(y-b)^2} \leq s-r$ and $(a,b,s)R(x,y,r)$ implies $\sqrt{(a-x)^2+(b-y)^2} \leq r-s$
After some algebra, here's what I get: For the first one, $x^2+y^2-s^2-2xa-2yb+2sr+b^2+a^2 \leq r^2$ and for the second one, $x^2+y^2-s^2-2ax-2by+2sr+b^2+a^2 \leq r^2$
and so they come out to be the same, but in an example from the notes, I think I should have came up with something like: $a \leq c$ and $c \leq a$, so $c=a$, and I didnt get anything of that form, so I'm wondering if my steps above are also correct. Have I proven anything, or if not, what have I done/assumed wrong?
Part (a) would give you the intuition. But the first thing to notice here (even before you notice that the square roots have the same value) is that since the square root is always nonnegative, the first condition gives you that $s-r\geq 0$ and the second gives you that $r-s\geq 0$. Those two together tell you what about $s$ and $r$?
Once you have that, you know something about the square root; and since the quantity inside is a sum of squares, the only way that something can happen is if both squares are -fill in the blank-. Which tells you something about $x$, $y$, $a$, and $b$.