Yesterday I needed to do some calculations with circles and "ventured" to calculate the arc length via the $\int{\sqrt{1+\left(f'(x)\right)^2}}$ formula and was baffled to see that in the case of unit half-circles that amounts to $\ ds := \sqrt{1+\left(f'(x)\right)^2} = \frac{1}{f(x)}$

Stuffing that differential equation in WA yields the four solutions $f(x) = \pm\sqrt{-2 c_1 x - c_1^2 - x^2 + 1}\ $and $f(x) = \pm\sqrt{2 c_1 x - c_1^2 - x^2 + 1}\ $

Questions:

has that special property been noted before?

are there other functions, whose arclength differential satisfies $\sqrt{1+\left(f'(x)\right)^2} = g\left(f(x)\right)$, when $g()$ is an ordinary, explicit function?