Answer
$(x-3)^2+(y-2)^2=4$
Work Step by Step
A circle has the equation:
$(x-h)^2+(y-k)^2=r^2$ with center $(h,k)$ and radius $r$
We are told that our circle has center $(3,2)$ (so $h=3$,$k=2$) and is tangent to the $x$-axis. If it is tangent to the $x$ axis, it must touch at $(3,0)$ (2 units below the center). Thus the radius must be $2$. Hence the equation is:
$(x-3)^2+(y-2)^2=2^2$
$(x-3)^2+(y-2)^2=4$