The tangent line at $(2,1)$ has slope $1$, so the radius to the point of tangency must be perpendicular to that line and therefore have slope $-1$. Since the circle passes through $(2,1)$ and $(8,1)$, its center must also lie on the perpendicular bisector of the segment joining those points. The midpoint of $(2,1)$ and $(8,1)$ is $(5,1)$, and because the segment is horizontal, its perpendicular bisector is the vertical line $x=5$. The radius line through $(2,1)$ with slope $-1$ has equation $y-1=-1(x-2)$, or $y=-x+3$. Intersecting this line with $x=5$ gives $y=-2$, so the center is $(5,-2)$. The radius is the distance from $(5,-2)$ to $(2,1)$:

Therefore, the statement that must be true is that the circle has radius $3\sqrt{2}$.

The distance around a circle is its circumference, given by the formula $C=2\pi r$. The radius of the splash pad is 6 feet, so substitute $r=6$ into the formula: $C=2\pi (6)=12\pi$. Therefore, the correct answer is $12\pi$.

A circle with center $(h,k)$ has equation $(x-h{)}^2+(y-k{)}^2=r^2$. Here, the center is $(4,-1)$, so the equation must begin as $(x-4{)}^2+(y+1{)}^2=r^2$. To find $r^2$, use the point $(10,7)$ on the circle:

Therefore, the boundary of the garden is $(x-4{)}^2+(y+1{)}^2=100$.

Because the circular path passes through $(2,6)$ and $(8,6)$, the center must lie on the perpendicular bisector of the segment joining those points. The midpoint of the segment is

Since the segment is horizontal, its perpendicular bisector is the vertical line $x=5$. The center is also given to lie on $y=2x-6$, so substitute $x=5$ into that line:

Therefore, the center is $(5,4)$. The radius is the distance from $(5,4)$ to either given point, such as $(2,6)$:

So the equation is

which is choice A.

The standard form of a circle with center $(h,k)$ and radius $r$ is $(x-h{)}^2+(y-k{)}^2=r^2$. Here, the center is $(3,-2)$, so $h=3$ and $k=-2$. Substituting gives $(x-3{)}^2+(y-(-2){)}^2=4^2$, which simplifies to $(x-3{)}^2+(y+2{)}^2=16$. Therefore, choice A is correct.