## College Algebra 7th Edition

$(x-7)^2+(y+3)^2=9$
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: $(x-7)^2+(y+3)^2=3^2 (x-7)^2+(y+3)^2=9$