#### Answer

$\color{blue}{(x-3)^2+(y+6)^2=36}$

#### Work Step by Step

RECALL:
The center-radius form of a circle whose center is at $(h, k)$ and whose radius is $r$ units is:
$(x-h)^2+(y-k)^2=r^2$
The given circle has its center at $(3, -6)$
Thus, the tentative equation of the circle is :
$(x-3)^2+(y-(-6))^2=r^2
\\(x-3)^2+(y+6)^2=r^2$
Since the circle is tangent to the x-axis, then the point on the x-axis directly above the center $(3, -6)$ must be a point on the circle. This point is $(3, 0)$.
To find the value of $r^2$, substitute the x and y values of the point $(3, 0)$ to obtain:
$(x-3)^2+(y+6)^2=r^2
\\(3-3)^2+(0+6)^2=r^2
\\0^2+6^2=r^2
\\0+36=r^2
\\36=r^2$
Therefore, the equation of the circle is:
$\\\color{blue}{(x-3)^2+(y+6)^2=36}$