To find the center and radius of the circle x^2+y^2-6x+2y-38>0, we need to first rewrite the equation in standard form by completing the square.
x^2+y^2-6x+2y-38>0
(x^2-6x)+(y^2+2y)-38>0
(x-3)^2-9+(y+1)^2-1-38>0
(x-3)^2+(y+1)^2>48
Now we can see that the equation is in the form (x-a)^2+(y-b)^2=r^2, where (a,b) is the center of the circle and r is the radius.
So the center of the circle is (3,-1) and the radius is sqrt(48), which simplifies to 4sqrt(3).