## Precalculus (10th Edition)

$(x-2)^2+(y-3)^2=3^2$.
The general equation for a circle with radius $r$ and centre $(h,k)$ is: $(x-h)^2+(y-k)^2=r^2$. The distance formula from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. If the circle is tangent to the x-axis, its radius is the absolute value of is y-coordinate, hence here: $r=3.$ Also $C=(2,3)$, hence our equation: $(x-2)^2+(y-3)^2=3^2$.