The orientated area $A(\gamma)$ of a regular closed plane curve $(\gamma, \tau)$ is defined as $$A(\gamma) :=\frac{1}{2}\int_{0}^\tau \det (\gamma,\gamma')$$ Now how can I determine the cricital points of the area functional $$\mathcal A:\mathcal M_L \to \Bbb R, \gamma \mapsto \mathcal A(\gamma)$$ where $\mathcal M_L$ denotes the space of closed plane curves $(\gamma,\tau)$ with length $L \gt0$, i.e. $\mathcal M_L = \{(\gamma,\tau) \vert \int_0^\tau\vert\gamma'\vert=L\}$.
Any hints on how to start would be much appreciated. Or maybe just an explanation on how I have to interpret $\det (\gamma,\gamma')$.
$\det(a,b) = a^TJb$ where $J = \left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]$.
Then the Euler-Lagrange equations of $$\frac{1}{2}\int_0^\tau \gamma^TJ\gamma' + \lambda(\|\gamma'\|-L/\tau)$$ are $$J\gamma' - \frac{d}{d\tau}\left(-J\gamma + \lambda \frac{\gamma'}{\|\gamma'\|}\right)=0$$ $$2J\gamma' - \frac{\lambda}{\|\gamma'\|}(I-TT^T)\gamma''=0$$ where $T=\frac{\gamma'}{\|\gamma'\|}$.
Multiplying both sides by $J\gamma'$ gives $$2\|\gamma'\|^2 = \frac{\lambda}{\|\gamma'\|} J\gamma' \cdot \gamma''$$ or $$\kappa = \frac{J\gamma' \cdot \gamma''}{\|\gamma'\|^3} = \frac{2}{\lambda}$$ and $\gamma$ has constant curvature, and so is a circle of length $L$.