SAT Question 76c73dbf: Circles answer and explanation with Desmos solution

It's given that the graph of $x^2+x+y^2+y=\frac{199}{2}$ in the xy-plane is a circle. The equation of a circle in the xy-plane can be written in the form ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where the coordinates of the center of the circle are $\left(h,k\right)$ and the length of the radius of the circle is $r$. The term ${\left(x-h\right)}^2$ in this equation can be obtained by adding the square of half the coefficient of $x$ to both sides of the given equation to complete the square. The coefficient of $x$ is $1$. Half the coefficient of $x$ is $\frac{1}{2}$. The square of half the coefficient of $x$ is $\frac{1}{4}$. Adding $\frac{1}{4}$ to each side of $\left(x^2+x\right)+\left(y^2+y\right)=\frac{199}{2}$ yields $\left(x^2+x+\frac{1}{4}\right)+\left(y^2+y\right)=\frac{199}{2}+\frac{1}{4}$, or ${\left(x+\frac{1}{2}\right)}^2+\left(y^2+y\right)=\frac{199}{2}+\frac{1}{4}$. Similarly, the term ${\left(y-k\right)}^2$ can be obtained by adding the square of half the coefficient of $y$ to both sides of this equation, which yields ${\left(x+\frac{1}{2}\right)}^2+\left(y^2+y+\frac{1}{4}\right)=\frac{199}{2}+\frac{1}{4}+\frac{1}{4}$, or ${\left(x+\frac{1}{2}\right)}^2+{\left(y+\frac{1}{2}\right)}^2=\frac{199}{2}+\frac{1}{4}+\frac{1}{4}$. This equation is equivalent to ${\left(x+\frac{1}{2}\right)}^2+{\left(y+\frac{1}{2}\right)}^2=100$, or ${\left(x+\frac{1}{2}\right)}^2+{\left(y+\frac{1}{2}\right)}^2=10^2$. Therefore, the length of the circle's radius is $10$.