$\DeclareMathOperator\LG{LG}$In the paper The - Conformal geometry of surfaces in the Lagrangian—Grassmannian and second order PDE (published on Proc. London Math. Soc.), I've found an interesting statement:

The Lie quadric $Q^3$, i.e., the space of all points, lines and circles of $\mathbb{R}^2$, is isomorphic to the Lagrangian Grassmannian $\LG(2,4)\subset \operatorname G(2,4)$ of all 2D subspaces of $\mathbb{R}^4$ which are isotropic with respect to the canonical symplectic form $\eta=e^1\wedge e^3+e^2\wedge e^4$, where $\{e^i\}$ is a dual basis of $\mathbb{R}^4$.

But this fact is not proved clearly, and I'm beginning to wonder whether it is actually true.

Just looking at the dimensions, everything looks all-right: $\LG(2,4)$ contains an open subset of symmetric $2\times 2$ matrices, so it must be 3D; to parameterise all circles in $\mathbb{R}^2$ (comprising zero and infinite radii) I need 3 numbers (2 for the position of the center and 1 for the radius).

Then, the actual proof should go as follows: Plücker-embed $\LG(2,4)$ into $\mathbb{P}(\ker \eta)\cong\mathbb{RP}^4$, where $\ker \eta\subset{\bigwedge}^2\mathbb{R}^4$, and observe that its image is the 3D quadric $Q\subset \mathbb{RP}^4$ of isotropic elements with respect to the obviously defined (conformal) symmetric form $$ {\bigwedge}^2\mathbb{R}^4\otimes{\bigwedge}^2\mathbb{R}^4\longrightarrow {\bigwedge}^4\mathbb{R}^4\cong\mathbb{R}. $$ On the other hand, $Q^3$ is the 3D quadric of isotropic lines in $\mathbb{R}^{2,3}$, but I cannot manage to make the last step and answer myself the following:

QUESTION:Is it true that $Q=Q^3$?

In the affirmative case, it should be such a basic fact that even several alternative proofs should have be given in literature—but I wasn't able to find any. For example, I'd like to see a very elementary one, i.e., not involving multilinear algebra, but just incidence/tangency of lines, planes and circles. Any reference will be appreciated!