The equation of a circle with center $(h,k)$ and radius $r$ is $(x-h{)}^2+(y-k{)}^2=r^2$. The center is $(3,-2)$, so the equation must begin as $(x-3{)}^2+(y+2{)}^2=r^2$. Since the circle passes through $(7,-2)$, the radius is the distance from $(3,-2)$ to $(7,-2)$, which is $7-3=4$. Therefore, $r=4$ and $r^2=16$. The correct equation is $(x-3{)}^2+(y+2{)}^2=16$.

A circle with center $(h,k)$ has equation $(x-h{)}^2+(y-k{)}^2=r^2$. Since the center is $(3,-2)$, the equation must begin as $(x-3{)}^2+(y+2{)}^2=r^2$. To find $r^2$, use the distance from the center to the point $(7,1)$:

So the equation is $(x-3{)}^2+(y+2{)}^2=25$.

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 $(4,-1)$, so the equation becomes $(x-4{)}^2+(y-(-1){)}^2=r^2$, which simplifies to $(x-4{)}^2+(y+1{)}^2=r^2$. The radius is 6, so $r^2=36$. Therefore, the boundary is represented by $(x-4{)}^2+(y+1{)}^2=36$.

Points $P(1,4)$ and $Q(5,4)$ have the same $y$-coordinate, so the segment $\overline{PQ}$ is horizontal. Its midpoint is $\left(\frac{1+5}{2},\frac{4+4}{2}\right)=(3,4)$. For a circle, the center lies on the perpendicular bisector of any chord, so $(3,4)$ is a strong candidate for the center. Now check the distance from $(3,4)$ to one of the given points: to $P(1,4)$, the distance is $2$, so the radius is $2$ and $r^2=4$. Therefore, the equation is $(x-3{)}^2+(y-4{)}^2=4$. As a quick verification, point $R(3,6)$ is also 2 units from $(3,4)$, so it fits the same circle.

The equation $(x-4{)}^2+(y+1{)}^2=25$ represents a circle with center $(4,-1)$ and radius $5$. To test choice A, find the distance from $(4,-1)$ to $(4,4)$. The horizontal change is $0$ and the vertical change is $5$, so the distance is $5$, which equals the radius. Therefore, $(4,4)$ lies on the circle, so this statement must be true.

You can also verify by substitution: $(4-4{)}^2+(4+1{)}^2=0+25=25$, which satisfies the equation.