Sides of Triangles Obtained
ABO	a/b + a	a +1	a/b +1
BCO	a + b	a + 1	b + 1
CDO	b + b/a	b + 1	b/a + 1
DEO	b/a + 1/a	b/a + 1	1/a + 1
EFA	1/a +1/b	1 + 1/b	1 + 1/a
FAO	a/b + 1/b	a/b + 1	1/b + 1

I observe that triangles

BCO

DEO

and

FAO

are similar, with the similarity ratios (taking them two by two),

1 : 1 / a : 1 / b

respectively. So are triangles

ABO

CDO

and

EFA

, with the similarity ratios

a : b : 1

Looking in the similarity ratio, I observe that angle $BOC$ is congruent with angle $DEO$ and with angle $AFO$ , angle $OBC$ with angles $FAO$ and $EOD$ , and angle $BCO$ with $ODE$ and $AOF$ .

This means the sum of angles $BOC$ , $AOF$ and $DOE$ is the angle sum in a triangle, i.e. $180^{0}$ .

For triangles $AOB$ , $COD$ and $EOF$ that are similar, angles $AOB$ , $ODC$ and $OEF$ are congruent; so are angles $OAB$ , $OFE$ and $DOC$ and $ABO$ , $OCD$ and $FOE$ . In this case the sum of angles: $AOB$ , $COD$ and $EOF$ is $180^{0}$ .

So, the sum of all angles around point O is $360^{0}$ . This means that, with the given radii, it is always possible to construct such a flower.