The standard form of a circle is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center. Since the center is $(4,-2)$, the equation must be $(x-4{)}^2+(y+2{)}^2=r^2$. The point $(1,-2)$ is 3 units from the center because the $x$-coordinates differ by $4-1=3$ and the $y$-coordinates are the same. So the radius is $3$, which means $r^2=9$. Therefore, the equation is $(x-4{)}^2+(y+2{)}^2=9$.

A circle with center $(h,k)$ has equation $(x-h{)}^2+(y-k{)}^2=r^2$. Since the center is $(2,-1)$, the equation must be $(x-2{)}^2+(y+1{)}^2=r^2$. To find $r^2$, use the point $(5,3)$ on the circle: $r^2=(5-2{)}^2+(3-(-1){)}^2=3^2+4^2=9+16=25$. So the equation is $(x-2{)}^2+(y+1{)}^2=25$.

The area of a circle is $A=\pi r^2$. Circle 1 has radius $4$, so its area is $\pi (4^2)=16\pi$. Circle 2 has area $\frac{9}{4}\cdot 16\pi =36\pi$. That means $\pi r^2=36\pi$, so $r^2=36$ and $r=6$. Since the center is still $(2,-1)$, the equation is $(x-2{)}^2+(y+1{)}^2=36$.

A circle in standard form is written as $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. Comparing $(x-4{)}^2+(y+1{)}^2=25$ to the standard form shows that the center is $(4,-1)$ and $r^2=25$. Therefore $r=5$, so the statement that must be true is that the circle has radius $5$.

The standard equation of a circle is $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. Here, the center is $(4,-1)$, so $h=4$ and $k=-1$. Substituting gives $(x-4{)}^2+(y-(-1){)}^2=6^2$. Simplifying, $(x-4{)}^2+(y+1{)}^2=36$. Therefore, the correct answer is A.