Finding a distance between a point in a circle from the center.

Assuming $\angle ABC=90^o$ is given.

You can get there slightly quicker:
By Pythagoras, $|AC|=\sqrt{40}$.
$OAC$ is isosceles, with $|OA|=|OC|=\sqrt{50}$.
You can then immediately get $\cos(\angle OAC)=\frac{|AC|/2}{|OA|} = \frac{\sqrt{40}/2}{\sqrt{50}}= \frac{1}{\sqrt{5}}$.
I don't yet see a way to shortcut the rest.

You could do it completely differently, by algebra. Use a coordinate system, centred on $B$, and let $O$ be the point $(x,y)$. Then we get two equations from the fact that $|OA|=|OC|=\sqrt{50}$.

$$x^2+(6-y)^2=50\\ (2-x)^2+y^2=50$$

There are fairly easily solved to give $y=1$, $x=-5$, from which you get $|OB|=\sqrt{26}$.

Refer to the figure:

$\hspace{4cm}$

From the right triangle $ACD$: $CD=\sqrt{AD^2-AC^2}=4\sqrt{10}$.

From similarity of right triangles $ABC$ and $CDE$: $$\frac{CE}{AB}=\frac{CD}{AC}\Rightarrow CE=12\\ DE=\sqrt{CD^2-CE^2}=4=BF\\ BE=CE-BC=12-2=10=DF=AF$$ Hence, $\angle DAF=45^\circ=\angle OAB$, indeed.

Finally, from the cosine theorem for $\triangle AOB$: $$\begin{align}BO&=\sqrt{AO^2+AB^2-2\cdot AO\cdot AB\cdot \cos \angle OAB}=\\ &=\sqrt{50+36-2\cdot \sqrt{50}\cdot 6\cdot \frac1{\sqrt2}}=\\ &=\sqrt{26}.\end{align}$$

A slight variation of the solution

Note that $R$ is circumradius of $\triangle ADC$,

\begin{align} |CD|&=2R\sin\alpha=2\sqrt5 ,\\ |BD|&=\sqrt{|CD|^2-a^2}=4 ,\\ |AD|&=c+BD=10 . \end{align}

By the Stewart’s Theorem for $\triangle AOD$,

\begin{align} |OD|^2\cdot c+|OA|^2\cdot |BD| -|AD|\cdot(|OB|^2+c\cdot |BD|) &=0 ,\\ |AD|\cdot ( R^2 -|OB|^2-c\cdot |BD|) &=0 , \end{align}

\begin{align} |OB|^2&= R^2-c\cdot |BD| \\ &=50-6\cdot4 =26 . \end{align}