You must memorize the standard equation of a circle 
with center = (h,k) and radius = r

(x - h)² + (y - k)² = r²

In your problem center = (h,k) = (3,-1) and 
radius = r = 2

So substitute (3) for h, (-1) for k and (2) for r.
DO NOT substitute anything for x or y, but leave them 
variable:

(x - (3))² + (y - (-1))² = (2)²

(x - 3)² + (y + 1)² = 4

That's the answer in standard form.  If you like you can
do some algebra on that and come up with the general form

x² + y² + Dx + Ey + F = 0

which will come out to be

x² + y² - 6x + 2y + 6 = 0

Edwin