= Solution
Let $\mathbf q=(x,y)$ be the centre displacement and $\theta$ the small disk rotation. Put $d=l-l_0>0$. The <moment of inertia of a uniform disc> is $I=ma^2/2$, obtained from $I=\int_0^a\rho r^2(2\pi r\,dr)$ and $m=\rho\pi a^2$. Thus the <kinetic energy> is $T=m|\dot{\mathbf q}|^2/2+ma^2\dot\theta^2/4$.
For one spring, let $\mathbf n_j$ point radially towards its fixed anchor and let $\mathbf t_j$ be the perpendicular unit <vector>. The moving attachment is $\mathbf q+aR(\theta)\mathbf n_j$, while the anchor is $(a+l)\mathbf n_j$. Expanding their separation to second order gives the spring extension beyond natural length:
$$
s_j-l_0=d-\mathbf n_j\cdot\mathbf q+\frac a2\theta^2+
\frac{(\mathbf t_j\cdot\mathbf q+a\theta)^2}{2l}+O((|\mathbf q|+|\theta|)^3).
$$
Insert this in the <elastic potential energy> $V=(k/2)\sum_j(s_j-l_0)^2$. For the three symmetric directions, $\sum_j\mathbf n_j=\sum_j\mathbf t_j=0$ and $\sum_j(\mathbf n_j\cdot\mathbf q)^2=\sum_j(\mathbf t_j\cdot\mathbf q)^2=3|\mathbf q|^2/2$. Dropping the constant equilibrium energy gives
$$
V_2=\frac{3k}{4}\left(1+\frac dl\right)(x^2+y^2)+\frac{3kda(a+l)}{2l}\theta^2.
$$
Therefore the quadratic <Lagrangian> is
$$
\boxed{\mathcal L_2=\frac m2(\dot x^2+\dot y^2)+\frac{ma^2}4\dot\theta^2-\frac{3k}{4}\left(1+\frac dl\right)(x^2+y^2)-\frac{3kda(a+l)}{2l}\theta^2.}
$$
The prestretch terms cannot be omitted: they supply transverse spring stiffness and the entire rotational restoring torque, as derived generally in <small oscillations of a disk attached to prestretched radial springs>.
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