(The following example avoids piecewise definitions; but there is nothing "special" about it.)
Consider the curve $\gamma$ with polar representation
$$r=r_0(\phi):=2+\cos(3\phi)\ ,$$
which looks like a clover leaf, see the following figure. This curve has a length $L_0$ and encloses an area $A_0$.

We now set up a perturbation of $\gamma$ in the form
$$r=r(\phi):=r_0(\phi)+ a + b\cos(3\phi)+c\cos\phi$$
with small parameters $a$, $b$, $c$. Define
$$L(a,b,c):=\int_0^{2\pi}\sqrt{r^2(\phi)+r'^2(\phi)}\ d\phi, \qquad A(a,b,c):={1\over2}\int_0^{2\pi}r^2(\phi)\>d\phi\ .$$
The quantities
$$a_{11}:={\partial L\over\partial a}\biggr|_{(0,0,0)},\quad a_{12}:={\partial L\over\partial b}\biggr|_{(0,0,0)},\quad a_{21}:={\partial A\over\partial a}\biggr|_{(0,0,0)},\quad a_{22}:={\partial A\over\partial b}\biggr|_{(0,0,0)}$$
are obtained by differentiation under the integral sign and putting $(a,b,c)=(0,0,0)$. Numerical integration then gives
$$a_{11}=4.4197,\quad a_{12}=9.3105\ ,$$
whereas the other two can be computed explicitly:
$$a_{21}=4\pi,\quad a_{22}=\pi\ .$$
At any rate, one has $\det[a_{ik}]\ne0$. From this we can draw the following conclusion:
The system of equations
$$L(a,b,c)=L_0, \quad A(a,b,c)=A_0$$
has the solution $(0,0,0)$. Therefore it can be solved for $a$ and $b$ in the neighborhood of $(0,0,0)$, which means that there are two $C^1$-functions
$c\mapsto \alpha(c)$, $c\mapsto\beta(c)$ with $\alpha(0)=\beta(0)=0$, such that
$$L\bigl(\alpha(c),\beta(c),c\bigr)=L_0,\quad A\bigl(\alpha(c),\beta(c),c\bigr)=A_0$$
for all $c$ in a neighborhood of $0$.
EDIT from String. This is such a great answer, so I decided that it should have an animation of its own:

I approximated $a$ and $b$ by iterating linearly solving the system of equations defined by the Jacobian. Three such linear steps were taken for each $c\in[-0.7,0.7]$ (by steps of $0.01$). For $c$'s of numerical value greater than about $1$ it appears that the system is still solvable, but the pertubation begins to intersect itself via loops so then it has no relevance to the problem in question.
@Christian Blatter: I hope you don't mind this edit - otherwise we can just delete it again!