Question 389908
a) The tangent lines are horizontal at the points (0,4) and (0,-4).

b) The tangent lines have slope 1 at the intersection points of the circle with the line y = -x.  Solving:
{{{x^2 + y^2 = x^2 + (-x)^2 = 2x^2 = 16}}} ==> {{{x^2 = 8}}}, or 
{{{x = 2sqrt(2)}}}, {{{x =  -2sqrt(2)}}}.
Then the points are ({{{2sqrt(2)}}},{{{ -2sqrt(2)}}}) and ({{{-2sqrt(2)}}},{{{ 2sqrt(2)}}}).