Question 768·Medium·Math

In standard form, $(x-h)^2+(y-k)^2=r^2$, the center is $(h,k)$. Here $x-3$ is zero at $x=3$, and $y+4$ is zero at $y=-4$, so the center is $(3,-4)$. The plus sign in $y+4$ does not mean the center’s $y$-coordinate is positive.

Right changes the center’s $x$-coordinate by $+2$; down changes its $y$-coordinate by $-5$. Apply those moves to the center, not to the numbers written inside the parentheses:

A translation changes position, not size, so keep $r^2=81$. Put the new center $(5,-9)$ into standard form:

Simplify the double negative:

Type the original and new equations on separate Desmos lines. Select each circle: Desmos marks their top points at $(3,5)$ and $(5,0)$. The second is $2$ right and $5$ down, as required. The new circle is $(x-5)^2+(y+9)^2=81$. Choice B.