SAT Question ee540927: Circles answer and explanation with Desmos solution

Choice D is correct. It's given that in the xy-plane, the graph of $x^2+58x+y^2=0$ is a circle. The equation of a circle in the xy-plane can be written as $(x-h{)}^2+(y-k{)}^2=r^2$, where the coordinates of the center of the circle are $(h,k)$ and the radius of the circle is $r$. By completing the square, the equation $x^2+58x+y^2=0$ can be rewritten as $\left(x^2+58x+{\left(\frac{58}{2}\right)}^2\right)+y^2=0+{\left(\frac{58}{2}\right)}^2$, or $\left(x^2+58x+841\right)+y^2=841$. This equation is equivalent to $(x+29{)}^2+y^2=841$, or ${\left(x-(-29)\right)}^2+(y-0{)}^2=841$. Therefore, $h$ is $-29$ and $k$ is $0$, and the coordinates $(x,y)$ of the center of the circle are $(-29,0)$.