Question

A particular circle in the standard (x,y) coordinate

Answer 1

For any circle with Cartesian equation
$(x-a)^2 + (y-b)^2 = r^2$,
we have that the centre of the circle is $(a,b)$, and the radius of the circle is $r$.

So in the case that 
$(x-5)^2 + y^2 = 38$,
we essentially have that
$a = 5, b = 0, r^2 = 38$.

So the centre of the circle is $(5,0)$, and the radius is $\sqrt{38}$.