TITLE: Constructing a family of distinct curves with identical area and perimeter
QUESTION [11 upvotes]: Two recent questions were posed by Arjuba [1] [2] asking for counter-examples regarding whether two different figures could have the same perimeter and area. Responders quickly raised a number of such examples. 
However, all were of the form of polygons rather than smooth curves. This doesn't pose any obvious conceptual obstacle, since one can certainly construct such a counter-example by a limiting process. But what I would be interested in seeing a construction for is a continuous family of smooth curves, all with identical area and perimeter. (A family of algebraic curves would be my preference). Are there well-known examples of such?
EDIT) To clarify my intent: What I really want is a continuous family of curves, either parameterized explicitly or given implicitly as an algebraic curve, which have the desired property.
EDIT 2) I've asked another question which suggests/ requests consultation on a possible variational strategy to finding examples with more smoothness.

REPLY [3 votes]: Take your favorite family of smooth Jordan curves parametrized by $\lambda>1$ say, for example $x^\lambda+y^\lambda=1$.  These all bound area between 2 and 4. Scale them to unit area by an appropriate factor depending on $\lambda$. The resulting curves don't have the same perimeter, but notice that each perimeter is less than 8, say. Now apply affine transformations of the form $\pmatrix{\mu & 0\\0 &\mu^{-1}}$. As $\mu$ tends to infinity the perimeter does also, so by the intermediate value theorem there is a $\mu=\mu(\lambda)$ making all perimeters equal to 8, yielding the desired family.
One can't do this with conic sections because there is not enough room, but cubic curves are enough; start for example with small perturbations of the circle: $x^2+y^2+\epsilon x^3=1$ with small $\epsilon$. These are algebraic and therefore analytic.
To make the examples more explicit, one needs to overcome the problem of the computation of arclength which typically leads to unmanageable integrals. Perhaps this is more tractable in polar coordinates where one can start with $r^2+\epsilon \sin\theta=1$ and hope for the best.