Pushing the boundaries: Codicil

Since $C$ is tangent to the $y$ -axis its radius will be $x$ . Let $P$ be the point $(0,y)$ , $Q(0,2)$ be the center of circle $A$ and $S(0,-1)$ be the center of circle $B$ .

Then $\Delta OPQ$ is a right triangle with legs length $x, (y - 2)$ and hypotenuse length $(x + 2)$ , and $\Delta OPS$ is a right triangle with legs length $x, (y + 1)$ and hypotenuse length $(x + 3)$ .

We thus have the following set of equations:

(i) $x^{2} + (y - 2)^{2} = (x + 2)^{2}$ and

(ii) $x^{2} + (y + 1)^{2} = (x + 3)^{2}$ .

After subtracting (i) from (ii) and simplifying we get $x = 3y - 4$ . Now substitute this back into (i) to find that

$(3y - 4)^{2} + (y - 2)^{2} = (3y - 2)^{2}$ ,

which after expanding and simplifying yields

$y^{2} - 16y + 16 = 0$ ,

which has solutions $y = \dfrac{16 \pm \sqrt{16^{2} - 4*16}}{2} = 8 \pm 4\sqrt{3}$ .

Now $y = 8 - 4\sqrt{3}$ implies that $x = 3y - 4 = 20 - 12\sqrt{3}$ , which is less than $0$ so can be discarded. That leaves us with

$y = 8 + 4\sqrt{3}$ and thus $x = 20 + 12\sqrt{3}$ .

Thus $x + y = 28 + 16\sqrt{3} = \boxed{55.713}$ to $3$ decimal places.