## College Algebra (11th Edition)

$(x-3)^2+(y-2)^2=4$
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: $(x-3)^2+(y-2)^2=2^2$ $(x-3)^2+(y-2)^2=4$