1. **State the problem:** Calculate the Radon transform of the function $$f(\mathbf{r}) = \exp \left[ - \frac{(\mathbf{r} - \mathbf{r}_0)^2}{2a^2} \right].$$ 2. **Recall the Radon transform definition:** The Radon transform $$g(s,\theta)$$ of $$f(\mathbf{r})$$ is given by integrating $$f$$ over the line parameterized by $$s$$ and direction unit vector $$\mathbf{n}_\theta$$: $$ g(s, \theta) = \int_{-\infty}^\infty f(s \mathbf{n}_\theta + t \mathbf{n}_\theta^\perp) dt, $$ where $$\mathbf{n}_\theta^\perp$$ is the unit vector perpendicular to $$\mathbf{n}_\theta$$. 3. **Substitute $$f$$ into the integral:** Since $$f(\mathbf{r})$$ is Gaussian centered at $$\mathbf{r}_0$$ with width $$a$$, $$ g(s, \theta) = \int_{-\infty}^\infty \exp \left[ - \frac{(s \mathbf{n}_\theta + t \mathbf{n}_\theta^\perp - \mathbf{r}_0)^2}{2a^2} \right] dt. $$ 4. **Simplify the quadratic form inside the exponent:** Write $$ (s \mathbf{n}_\theta + t \mathbf{n}_\theta^\perp - \mathbf{r}_0)^2 = (s - \mathbf{r}_0 \cdot \mathbf{n}_\theta)^2 + (t - \mathbf{r}_0 \cdot \mathbf{n}_\theta^\perp)^2 $$ because $$\mathbf{n}_\theta$$ and $$\mathbf{n}_\theta^\perp$$ are orthonormal. 5. **Integrate over $$t$$:** The integral becomes $$ g(s, \theta) = \exp \left[ - \frac{(s - \mathbf{r}_0 \cdot \mathbf{n}_\theta)^2}{2a^2} \right] \int_{-\infty}^\infty \exp \left[ - \frac{(t - \mathbf{r}_0 \cdot \mathbf{n}_\theta^\perp)^2}{2a^2} \right] dt. $$ 6. **Use the Gaussian integral formula:** $$ \int_{-\infty}^\infty \exp \left[-\frac{(t - \mu)^2}{2a^2} \right] dt = a \sqrt{2 \pi} $$ for any mean $$\mu$$. 7. **Combine results:** Thus, $$ g(s, \theta) = \sqrt{2\pi} a \exp \left[ - \frac{(s - \mathbf{r}_0 \cdot \mathbf{n}_\theta)^2}{2a^2} \right]. $$ **Final answer:** $$ g(s, \theta) = \sqrt{2\pi} a \exp \left[ - \frac{(s - \mathbf{r}_0 \cdot \mathbf{n}_\theta)^2}{2a^2} \right].$$