Exponential fades are not circles, but they do look a bit like ellipse quadrants. If that's not what you mean, you'll have to find a picture (or look here). If it is, here you go: Exponential curves on linear impulses (not dB) sound smooth and straight, because the ear translates takes a logarithm of them (as it does with frequencies) to become linear. This is called an exponential fade. By changing the base you can affect the sharpness of the fade.
Generally you want to avoid finding matches on a curve for your time & amplitude, so you'll stretch the exponential's known points to your start and end points and amplitude. Exponentials of all bases b go through the points (0,1) and (1,b), and we want them to go through (start, volstart) and (end, volend).
Crop the output by subtracting 1 from the whole function, so that b^time becomes b^time-1 (not b^(time-1) ), and those points become (0,0) and (1,b-1). Stretch the amplitude by dividing by b-1 and multiplying by volend-volstart, sending (1,b-1) to (1,1) then (1,volend-volstart). The function is now: (b^time-1)*(volend-volstart)/(b-1). Raise this clip to meet your desired amplitudes by adding volstart, so that the points are now (0,volend), (1,volend), and the function is (b^time-1)*(volend-volstart)/(b-1)+volstart.
All that's left now is stretching and clipping the input, which can be done a number of ways, but all entail turning start-end into 0-1. You can give this expression a time input from a [line] that ramps from 0 to 1, from start until end in ms. You could, of course, run that line through another function first, which still outputs from 0 to 1, and make a more complex envelope. You can bake the stretching and clipping into the expression by turning time into (time-start)/(end-start), so that as time runs from start to end, the function outputs as if running from 0 to 1.