Circle in a Quadrant

$$ See the picture below:

Circle in Circle

We can obtain $\overrightarrow{AO}=r\sqrt{2}$ , $\;\overrightarrow{OB}=r$ , and $\;\overrightarrow{AB}=R$ . Therefore

$\begin{aligned} \overrightarrow{AB}&=\overrightarrow{AO}+\overrightarrow{OB}\\ R&=r\sqrt{2}+r\\ R&=r(\sqrt{2}+1)\\ r&=\frac{R}{1+\sqrt{2}}\\ &=\frac{R}{1+\sqrt{2}}\cdot\frac{1-\sqrt{2}}{1-\sqrt{2}}\\ &=\frac{R(1-\sqrt{2})}{1-2}\\ &=(\sqrt{2}-1)R. \end{aligned}$

For $R=1$ unit, we obtain $r=(\sqrt{2}-1)$ unit.

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$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

The quadratic eqn for radius of inner circle (r) would work out to

r^2 + 2r -1 = 0 .. solving for roots -1+sqrt(2) is the positive root.