Answer
$(x-2)^2+(y-3)^2=9$
Work Step by Step
RECALL:
The standard equation of a circle whose center is at the point $(h, k)$ is:
$(x-h)^2+(y-k)^2=r^2$
where $r$ = radius
With its center at $(2, 3)$, the tentative equation of the given circle is:
$(x-2)^2+(y-3)^2=r^2$
Since the circle is tangent to the $x$-axis, the point on the x-axis that is directly below the center is a point on the circle. This point is $(2, 0)$.
Note that the distance from the center $(2, 3)$ to the point on the circle $(2, 0)$ is 3 units.
Therefore, the standard form of the equation of the given circle is:
$(x-2)^2+(y-3)^2=3^2
\\(x-2)^2+(y-3)^2=9$
