The question is from the 10th page of Topping's Lectures on Ricci flow, the calculation about Hamilton's cigar soliton.
Consider $R^2$ with metric $$ g=\rho^2 (dx^2 + dy^2),~~~\rho^2= \frac{1}{1+x^2 + y^2} $$ where $dx^2 = dx \otimes dx $. Topping state that $$ g=ds^2 + \tanh^2 s d\theta^2 $$ in terms of the geodesic distance from the origin $s$, and the polar angle $\theta$. But I can't get it.
What I try: In polar coordinates $$ x=r\cos\theta,~~~ y = r\sin\theta $$ I get $$ g=\rho^2 dr^2 + r^2 \rho^2 d\theta^2 $$ The process: $$ \begin{pmatrix} dx \\ dy \end{pmatrix} = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix} \begin{pmatrix} dr \\ d\theta \end{pmatrix} $$ and \begin{align} g &= \begin{pmatrix} dx &dy \end{pmatrix} \begin{pmatrix} \rho^2 & 0 \\ 0 & \rho^2 \end{pmatrix} \begin{pmatrix} dx \\ dy \end{pmatrix} \\ &= \begin{pmatrix} dr &d\theta \end{pmatrix} \begin{pmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{pmatrix} \begin{pmatrix} \rho^2 & 0 \\ 0 & \rho^2 \end{pmatrix} \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix} \begin{pmatrix} dr \\ d\theta \end{pmatrix} \\ &= \rho^2 dr^2 + r^2 \rho^2 d\theta^2 \end{align} but I don't know how to deal the geodesic distance $s$.
As you noted, in polar coordinates, the metric reads $$ g = \frac{1}{1+r^2}\mathrm{d}r^2 + \frac{r^2}{1+r^2}\mathrm{d}\theta^2 $$ Heuristically, you are looking for $s$ such that $\mathrm{d}s^2 = \frac{1}{1+r^2}\mathrm{d}r^2$, which can be stated as ${\mathrm{d}s} = \frac{1}{\sqrt{1+r^2}}{\mathrm{d}r}$. So let's define $$ s(r) = \int_0^r \frac{1}{\sqrt{1+t^2}}\mathrm{d}t = \sinh^{-1}(r) $$ It now follows that $$ \frac{r^2}{1+r^2} = \frac{\sinh(s)^2}{1+\sinh(s)^2} = \frac{\sinh(s)^2}{\cosh(s)^2} = \tanh(s)^2 $$ and finally, $$ g = \mathrm{d}s^2 + \tanh(s)^2\mathrm{d}\theta^2 $$