# Relations Problem 3

Reminder of how we defined the relation:

(x,y,r) R (a,b,s) if and only if $$\sqrt{(x-a)^2 + (y-b)^2} \le s- r$$

### Solution to (a)

Notice that $$q = \sqrt{(x-a)^2 + (y-b)^2}$$ is the distance between the centers of the two circles. We can rephrase $$\sqrt{(x-a)^2 + (y-b)^2} \le s- r$$ as $$r + \sqrt{(x-a)^2 + (y-b)^2} \le s$$. Think about measuring outwards from the center of the larger circle. Then $$r + \sqrt{(x-a)^2 + (y-b)^2}$$ is the distance to the center of the smaller circle, plus the radius of the smaller circle, i.e. the distance to the far side of the smaller circle. This is no bigger than s, which is the distance to the edge of the larger circle. In other words, the smaller circle (x,y,r) is contained inside the larger circle (a,b,s).