$|x+y| = R(2-\sqrt{2})$

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How did you get to $135^{\circ}$ on $P$? I'm still confused on what sort of relation does it exist between $R$ and $\vec{y}$? I'm lost at geometry. Can you add the justification for the $R$? Please, I'm still stuck on that part.

The only thing which it comes to my mind is that a ray between the center of the quarter of the circle connects it to $P$ and this ray divides the right angle in two equal halves $45^{\circ}$ and $45^{\circ}$, but why? What's the geometrical justification for this?. Given this condition, I'm assuming that this also repeats for $\vec{x}$, which makes it the same size (or norm) for $\vec{y}$.

In any case I'm getting:

$\left\|\vec{x}\right\|=\left\|\vec{y}\right\|=R\sqrt{2-\sqrt{2}}$

Then repeating the cosines law I'm getting:

$\left\|x+y\right\|^2=\left(R\sqrt{2-\sqrt{2}}\right)^2+\left(R\sqrt{2-\sqrt{2}}\right)^2-2\left(R\sqrt{2-\sqrt{2}}\right)^2\cos 45^{\circ}$

$\left\|x+y\right\|^2=2(R^2)(2-\sqrt{2})\left(1-\frac{\sqrt{2}}{2}\right)$

$\left\|x+y\right\|^2=R(2-\sqrt{2})$

This is the only method which I've found to get to the answer, but I'd like to know a geometrical justification for this. Please.