Waste your time on this, best done drunk.
Imagine a line with points A and B 980mm apart with a circle’s circumference touching both points. At a position mid way between AB the circle’s circumference must be 40mm perpendicular distance from the line AB. What is the radius of the circle?
It’s for the woodwork, give it a fucking go and help me out 
Just when did we go metric, or have I missed something?
Don’t worry, we will go back to proper measurements, rods, poles and perches, chains, furlongs etc.
Brexit will make it happen
The answer is you are going to need a bigger workbench.
By Pythagoras I think it’s root (49^2+45^2)
Which is 665mm
10ft (near enough for woodwork).
The trick is to spot that the triangle between the point A, the centre of the circle (let’s call that O) and the point where your 40mm perpendicular springs from (let’s call that C) is a right-angled triangle. Then Pythagoras will do the job for you. Since the distance from the circle’s centre, O, to any point on it is R (the radius) the distance OC must be R-40mm. Here you go:
EDIT: I guess it’s obvious, but my drawing is very much not to scale.
VB
About 10ft.
Oops… I see Graeme got there first.
I thought I’d better check. It turns out that if you use 10ft your circle will project 39.64mm instead of 40mm. Given the tightness of Bob’s immaculate jointing perhaps a third of a millimetre will matter actually ?
EDIT: You’ll get back to a 40mm projection if you make the distance AB not 980mm but 38.75 inches though. Which it is, near enough.
VB
thanks very much all. I shall make a template on the living room floor when SWMBO is out. I can tap a nail in, use a piece of string and a pencil and mark it out on a thin sheet of MDF. Will then transfer this to the workpiece.
Cheers all.
