I think that a straight variational approach has a good chance of yielding the desired conclusion. Fix two points $z_0,z_1$ on the boundary of the unit disk. Denote by $C$ the positively oriented arc of the circle that runs from $z_0$ to $z_1$.

Let $\mathcal{P}_{z_0,z_1}$ the set of paths $[0,1]\to \mathbb{C}$ with endpoints $z_0$ $z_1$. $\mathcal{P}_{z_0,z_1}$ is an affine space modeled on the vector space of maps $[0,1]\to\mathbb{C}$ that vanish at endpoints.

If $\gamma\in \mathcal{P}_{z_0,z_1}$ is an *embedded* path *inside* the disk, then the area between $\gamma$ and $C$ is given by the integral

$$\int_C xdy -\int_\gamma xdy $$

Your constraint can now be given a simpler form

$$\int_\gamma xdy =const. $$

Now solve the constrained variational problem

$$ \min\left\lbrace \int_0^1 |\dot{\gamma}(t)| dt;\;\;\gamma\in \mathcal{P}_{z_0,z_1},\;\;\int_\gamma xdy =const\right\rbrace. $$

If we write $\gamma(t)= x(t) + i y(t)$ then the constraint equation can be rewritten as

$$ \int_0^1 x(t) \dot{y}(t) dt =const. $$

This defines a quadratic hypersurface in the affine space $\mathcal{P}_{z_0,z_1}$.

If we define

$$L, F: \mathcal{P}_{z_0,z_1}\to \mathbb{R}, $$
$$ L(\gamma)= \int_0^1 |\dot{\gamma}(t)| dt,\;\; F(\gamma)= \int_\gamma xdy$$

then the **Euler-Lagrange** equations for the above variational problem have the form

$$ dL +\lambda dG=0 \tag{$EL_\lambda$}$$

where $\lambda\in\mathbb{R}$ is a Lagrange multiplier and $dL$ and $dF$ are the differentials of $L$ resp. $F$. The differential of $F$ is a linear function so ($EL_\lambda$) can be viewed as a nonlinear eigenvalue problem. It can be written very explicitly and I believe that playing with it will yield the desired conclusion.