I denoted by $A$ the centre of the circle of radius $a/b$, by $B$ the centre of the radius $a$, by $C$ the circle of radius $b$, by $D$ the centre of the circle of radius $b/a$, by $E$ the centre of the circle of radius $1$/a, by $F$ the centre of the circle of radius $1/b$ and by $O$ the centre of the circle of radius 1.

I observed that the triangles obtained have the following sides:

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.