Eight years late, but never mind; this is a really interesting problem that has popped up at random. Jer, I’m with you all the way in your 1st post, and agree with your numeric answers in your 2nd post. My differentiation to minimise equation [7] gave
3(b2 – 1) = b2, leading to b = sqrt(3/2) = 1.2247.. (as you found numerically).
For part 2, I wondered if we could fit the three circles in a line, touching each other, by using d = 2, but this led to a circumscribing ellipse that was bigger than your circular solution. Maybe you had tried that
Then I wondered if we could nudge the middle circle slightly out of line so that d is reduced. It sounds like a complication, but in fact the maths gets easier: Using Pythagoras in the triangle with vertices at (0,0), (d,0) and the centre of the ‘middle’ circle (which is in its extreme position, touching the ellipse at (0,b)):
gives:d2 = 4 - (b – 1)2.Substituting this in your equation [5] and simplifying
then gives:a = b*sqrt(2/(b – 1)), so thata*b = b2*sqrt(2/(b – 1)).
Using differentiation to minimise a*b now gives b = 4/3,
soa = (4/3)*sqrt 6,d = (1/3)*sqrt(35)and the area is 13.6805..,
which is slightly smaller than that of your circumscribing circle.
Belated thoughts (spoiler)
