Question

On the figure given show that `bar(OC)` is `sqrt(2)`?

Answer 1

WOW...I finally got it...although it seems too easy...and probably it is not the way you wanted it!

Explanation:

I considered the two small circles as equal and having radius `1`, each of them (or `u` as unity in distance `bar(PO)`...I think). So the entire base of the triangle (diameter of big circle) should be `3`.
According to this, the distance `bar(OM)` should be `0.5` and the distance `bar(MC)` should be one big cirlce radius or `3/2=1.5`.
Now, I applied Pythagoras to the triangle `OMC` with:
`bar(OC)=x`
`bar(OM)=0.5`
`bar(MC)=1.5` and I got:
`1.5^2=x^2+0.5^2`
or:
`x^2=1.5^2-0.5^2=(3/2)^2-(1/2)^2=8/4=2`
so:
`x=sqrt(2)`

Does it make sense...?