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  1. #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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