1 | initial version |

The above is a more or less analytic answer. It's possible (perhaps not advisable) to give one from the data itself, though this is probably not robust.

```
sage: P = parametric_plot((ff*cos(t),ff*sin(t)), (t,0,2*pi) , fill=True, fillcolor='orange', figsize=[3,3])
sage: p = P[0]sage: Z = zip(p.xdata,p.ydata)
sage: W = [(vector(z).norm(),z) for z in Z]
sage: W.sort()
sage: W[-5:]
[(4.07480714946, (3.0424694670479733, -2.7106148467446545)), (4.0760691956, (2.9829635305007063, -2.7778172479461185)), (4.07657737686, (3.028814532960587, -2.7285097827236444)), (4.07721170912, (2.99905277507801, -2.762125589698095)), (4.07738007646, (3.01433945952578, -2.7456849620178922))]
```

The last two points correspond to the outer parts of the biggest leaves in your picture. Note that this involves a lot of rounding error, relatively speaking. Using

```
sage: P = parametric_plot((ff*cos(t),ff*sin(t)), (t,0,2*pi) , fill=True, fillcolor='orange', plot_points=400)
```

gives things slightly closer together.

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