Radius Ratio #3

Let $C_{2}$ have equation $(x - b)^{2} + (y - b)^{2} = b^{2}$ . Then since the center $(a, a)$ of $C_{1}$ lies on $C_{2}$ we have that

$(a - b)^{2} + (a - b)^{2} = b^{2} \Longrightarrow (\frac{a}{b} - 1)^{2} = \frac{1}{2} \Longrightarrow \frac{a}{b} = 1 \pm \frac{1}{\sqrt{2}}$ .

Now since $b \gt a$ we must have that $\frac{a}{b} = 1 - \frac{1}{\sqrt{2}}$ , and thus

$\frac{b}{a} = \dfrac{\sqrt{2}}{\sqrt{2} - 1} = 2 + \sqrt{2}$ .

Therefore $x = 2, y = 2$ and $x + y = \boxed{4}$ .

Imgur

Taking a = 1, and b = R as shown

from similar triangles OC1M and OC2N $\frac{R + \sqrt{2}}{\sqrt {2}} = \frac{R}{1}$ $\frac{b}{a} = R = \frac{\sqrt{2}}{\sqrt{2}-1} = 1 + \sqrt{2}$