Let $S^1=\{\omega\in\mathbb{R}^2:|\omega|=1\}$ be the unit circle, let $\mathbb{R}_0=\mathbb{R}\setminus\{0\}$, and let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Let $R: C_c^\infty(\mathbb{R}^2) \to \mathcal{D}'(S^1 \times \mathbb{R})$ be the plane Radon transform, with operator convention determined by the oscillatory kernel phase

\begin{align*} \phi: S^1 \times \mathbb{R} \times \mathbb{R}^2 \times \mathbb{R} \to \mathbb{R}, \quad \phi(\omega,s,x,\lambda)=\lambda(x\cdot \omega-s). \end{align*}

Using the Fourier integral operator convention that an oscillatory kernel with phase $\phi(y,x,\lambda)$ has canonical relation

\begin{align*} C=\{(y;d_y\phi),(x;-d_x\phi): \partial_\lambda \phi=0,\ \lambda \ne 0\}, \end{align*}

the plane Radon transform is associated with the homogeneous canonical relation

\begin{align*} C_R=\{((\omega,s;\eta_\omega,\eta_s),(x;\xi))\in T^*(S^1\times \mathbb{R})\times T^*\mathbb{R}^2 : x\cdot \omega=s,\ \xi=-\lambda\omega,\ \eta_s=-\lambda,\ \eta_\omega=\lambda\,x|_{T_\omega S^1},\ \lambda\in \mathbb{R}_0\}. \end{align*}

Here $x|_{T_\omega S^1}\in T_\omega^*S^1$ denotes the covector $v\mapsto x\cdot v$ on $T_\omega S^1$.