"From the endpoints of diameter OO′of the outer circle we construct, on the same side, two tangents to the inner circle. Prove they are perpendicular.
I think
Since the base of the triangle is the diameter the angle would be worth $90^o$ . But since it would be a very simple solution, he put the angle outside the circumference. I believe that the idea here would be to show that the tangents are perpendicular to each other and therefore $x$ would be worth $90^o$ and belong to the circumference. But I couldn't do this demosntration
To help with the simplified resolution that was posted.
$\triangle YO'P: YP^2 = (R+x)^2 - r^2$
T.Fuss: $\dfrac{R+x}{R-x} = \dfrac{\sqrt{(R+x)^2-r^2}}r = \dfrac{YP}{r} = cotg(\angle XYZ) = \frac1{\dfrac{R-x}{R+x}} = tg (\angle XZY)$
$\triangle ZQO': ZQ^2 + r^2 = (R-x)^2 \implies ZQ = \sqrt{(R-x)^2 - r^2}$
T.Fuss: $\dfrac{R-x}{R+x} = \dfrac{\sqrt{(R-x)^2- r^2}}r = \dfrac{ZQ}r = cotg(\angle XZY)$
$\therefore \angle XZY + \angle XYZ = 90^{\circ} \implies \angle ZXY = 90^{\circ}$
(Credits: FelipeMartin)
Let $r,R,X$ be the inradius, circumradius and $OO'$.
By Fuss' Theorem for bicentric quadrilaterals $$\frac{1}{(R-X)^2}+\frac{1}{(R+X)^2}=\frac{1}{r^2}.$$ This can be rewritten as $$\frac{R+X}{R-X}=\frac{\sqrt{(R+X)^2-r^2}}{r}.$$
In the triangle containing angle $x$ draw in the lines of length $r$ from $O'$ to the opposite sides of the triangle. This produces two right-angled triangles and the above equation shows that these are similar.
This immediately gives $x=90^0$.