sat-dog

Choice D is correct. The graph in the xy-plane of an equation of the form ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ is a circle with center $\left(h,k\right)$ and a radius of length $r$. It's given that circle A is represented by $x^2+{\left(y-1\right)}^2=49$, which can be rewritten as $x^2+{\left(y-1\right)}^2=7^2$. Therefore, circle A has center $\left(0,1\right)$ and a radius of length $7$. Shifting circle A down two units is a rigid vertical translation of circle A that does not change its size or shape. Since circle B is obtained by shifting circle A down two units, it follows that circle B has the same radius as circle A, and for each point $\left(x,y\right)$ on circle A, the point $\left(x,y-2\right)$ lies on circle B. Moreover, if $\left(h,k\right)$ is the center of circle A, then $\left(h,k-2\right)$ is the center of circle B. Therefore, circle B has a radius of $7$ and the center of circle B is $\left(0,1-2\right)$, or $\left(0,-1\right)$. Thus, circle B can be represented by the equation $x^2+{\left(y+1\right)}^2=7^2$, or $x^2+{\left(y+1\right)}^2=49$.

Choice A is incorrect. This is the equation of a circle obtained by shifting circle A right $2$ units.

Choice B is incorrect. This is the equation of a circle obtained by shifting circle A up $2$ units.

Choice C is incorrect. This is the equation of a circle obtained by shifting circle A left $2$ units.