I'm not sure that I understand the question. That function f(x,y) is always 1 except at the origin. I don't see what that has to do with a function that will return 1 if a point is in the unit circle and 0 otherwise.

You can define the given function f(x,y) in terms of the hardlim function by

f(x,y) = hardlim( hardlim(x) + hardlim((–1)*x) + hardlim(y) + hardlim((–1)*y) ).