Answer
$(x-7)^{2}+(y+3)^{2}=9$
Work Step by Step
Center $(7,-3);$ tangent to the $x$-axis
The equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is its radius.
The center of the circle and the fact that it is tangent to the $x$-axis are given.
Because the circle is tangent to the $x$-axis, it must pass through the point $(7,0)$. Find the distance between this point and the center of the circle to obtain its radius. The formula for the distance between two points is $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
$d=\sqrt{(7-7)^{2}+(0+3)^{2}}=\sqrt{0^{2}+3^{2}}=\sqrt{0+9}=3$
Since now both the radius and the center of the circle are known, substitute them into the equation of a circle formula:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
$(x-7)^{2}+(y+3)^{2}=9$
