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07-07-2019 #2
This one doesn't need much more than Pythagoras to solve it.
Call the distance of the centre of a blue circle from the horizontal line h (note - this is not the centre-centre spacing of the two blue circles)
Call the distance of the centre of a blue circle from the vertical line x
Call the radius of the small circle r
Then the radius of the large circle is: (h^2+x^2-r^2)/(2*(x-r))
I would have shown my working but typing out equations like this is too tedious!
No guarantees, but I've checked the results for a few values against a sketch in Fusion 360 and it gives the same answers. F360 won't let you construct a construct a circle with those constraints either, but you can create a circle in roughly the right position and then apply appropriate constraints afterwards to get the same effect.Last edited by Neale; 07-07-2019 at 09:37 PM.
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