Answer
$(x-7)^2+(y+3)^2=9$
Work Step by Step
RECALL:
The standard form of a circle whose center is at $(h, k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$.
The center is at $(7, -3)$, so $h=7$ and $k=-3$.
Thus, the standard form of the given circle's equation of the circle is:
$(x-7)^2+[y-(-3)]^2=r^2
\\(x-7)^2+(y+3)^2=r^2$
The circle is tangent to the x-axis.
This means that the circle touches the x-axis at the point directly below the center, which is the point $(7, 0)$
The circle passes through the point $(7, 0)$.
This point is directly below the center, and is actually 3 units away from the center.
This means that the radius of the circle is 3 units.
Therefore, the standard form of the given circle's equation is:
$(x-7)^2+(y+3)^2=3^2
(x-7)^2+(y+3)^2=9$
