Find the distance between the centre from a vertex of a inside triangle of a circle. 
A machine-shop cutting tool has the shape of a botched circle, as shown. The radius of the circle $\sqrt{50}$ 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 centre of the circle.

Please give me some for this problem.I failed to find any way to attack. If you give me the complete solution then it will help me more. Thank you.
 A: 
Calculate the length AC and the angle at $A$ in the first triangle.
Calculate the angle at $A$ in the second triangle, this will allow you to calculate the angle at $A$ in the third triangle.
Use the cosine rule on the third triangle.
A: The distance BO is $\sqrt{26}$. 
We can show this is dropping perpendiculars etc, but to simplify (and also because I cannot draw on my phone), we can do this:
Turn the figure almost upside down so that AC (of length $2\sqrt{10}$ ) becomes horizontal. There are two triangles sitting on top of AC. One is ABC and the other triangle is the isosceles triangle COA where O is the center of the circle. We want the distance OB between the third vertices of these triangles which share side AC. 
Let E be the point where the perpendicular from O meets AC. 
$AE=\sqrt{10}$ and by Pythagoras theorem,  $OE=2\sqrt{10}$. 
Drop a perpendicular from B to AC, which meets it at D. Note similarity of ADB to ABC whose 3 sides are known and immediately deduce sides AD and DB. 
Now use Pythagoras one final time 
$$OB^2=(AE-AD)^2+(OE-DB)^2=26$$ which gives us $OB=\sqrt{26}$
