For any circle with Cartesian equation
[tex](x-a)^2 + (y-b)^2 = r^2[/tex],
we have that the centre of the circle is [tex](a,b)[/tex], and the radius of the circle is [tex]r[/tex].
So in the case that
[tex](x-5)^2 + y^2 = 38[/tex],
we essentially have that
[tex]a = 5, b = 0, r^2 = 38[/tex].
So the centre of the circle is [tex](5,0)[/tex], and the radius is [tex]\sqrt{38}[/tex].