MHB Find the square of the distance

[anemone]

The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is 6 cm and that of $BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance from $B$ to the center of the circle.

[TIKZ]
\draw[purple,thick] (0,0) arc (70:328:3);
\coordinate[label=above:A] (A) at (0,0);
\coordinate[label=left:B] (B) at (0,-4.43);
\coordinate[label=right:C] (C) at (1.48,-4.43);
\draw[thick] (A) -- (B) -- (C);
[/TIKZ]

 

[Opalg]

Spoiler

[TIKZ][scale=1.3]
\draw[purple,thick] (0,0) arc (70:328:3);
\coordinate[label=above:$A$] (A) at (0,0);
\coordinate[label=left:$B$] (B) at (0,-4.43);
\coordinate[label=right:$C$] (C) at (1.48,-4.43);
\coordinate[label=left:$O$] (O) at (-1.03,-2.82);
\coordinate[label=right:$N$] (N) at (0.74,-2.21);
\draw[thick] (A) -- (B) -- (C);
\draw (C) -- (A) -- (O) -- (B) ;
\draw (N) -- (O) ;
\draw (0.12,-0.8) node{$\beta$} ;
\draw (-0.15,-0.8) node{$\alpha$} ;
[/TIKZ]
In the diagram, $O$ is the centre of the circle, $N$ is the foot of the perpendicular from $O$ to $AC$, and the angles $\alpha$, $\beta$ are as shown.

By Pythagoras, $AC = \sqrt{40}$, so $AN = \sqrt{10}$. By Pythagoras again, $ON = \sqrt{40}$. Then $\tan\beta = \frac13$ and $\tan(\alpha + \beta) = 2$. Therefore $$\tan\alpha = \frac{2-\frac13}{1 + \frac23} = 1$$ and so $\alpha = 45^\circ$.

Now use the cosine rule in triangle $OAB$ to get $OB^2 = 50 + 36 - 2*\sqrt{50}*6*\frac1{\sqrt2} = 86 - 60 = 26.$