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 $(x-4{)}^2+(y+1{)}^2=100$.

A circle with center $(h,k)$ has equation $(x-h{)}^2+(y-k{)}^2=r^2$. Since the fountain is at the center, the circle’s center is $(4,-1)$, so the equation must be of the form $(x-4{)}^2+(y+1{)}^2=r^2$. To find $r^2$, use the point $(10,7)$ on the circle:

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

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 have the form $(x-4{)}^2+(y+1{)}^2=r^2$. To find $r^2$, use the point $(8,2)$ on the circle. The horizontal change from the center is $8-4=4$, and the vertical change is $2-(-1)=3$. Thus, by the distance formula, $r^2=4^2+3^2=16+9=25$. Therefore the equation is $(x-4{)}^2+(y+1{)}^2=25$.

Because $PA$ is tangent to the circle at $A$, the radius $OA$ is perpendicular to $PA$. That means triangle $OPA$ is a right triangle with hypotenuse $OP=13$ and leg $OA=5$. Using the Pythagorean theorem:

So the length of $PA$ is $12$ feet.

The standard form of a circle's equation is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center. Since the center is $(4,-1)$, the equation must begin as $(x-4{)}^2+(y+1{)}^2=r^2$. Next, find $r^2$ using the point $(10,7)$ on the circle. The horizontal change from the center is $10-4=6$, and the vertical change is $7-(-1)=8$. So $r^2=6^2+8^2=36+64=100$. Therefore, the equation of the boundary is $(x-4{)}^2+(y+1{)}^2=100$.