find an equation of a circle that satisfies the given conditions through 3,4 slope 0

all we know is that the center of the circle lies on the line x=3. (why?) So, if its radius is r, we have

(x-3)^2 + (y-(4-r))^2 = r^2
So I guess the simplest case would be r=4, giving
(x-3)^2 + y^2 = 16