Circles SAT question

The correct answer is $10$. 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 ${(x-h)}^2+{(y-k)}^2=r^2$, where the coordinates of the center of the circle are $(h,k)$ and the length of the radius of the circle is $r$. The term ${(x-h)}^2$ 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 $(x^2+x)+(y^2+y)=\frac{199}{2}$ yields $(x^2+x+\frac{1}{4})+(y^2+y)=\frac{199}{2}+\frac{1}{4}$, or ${(x+\frac{1}{2})}^2+(y^2+y)=\frac{199}{2}+\frac{1}{4}$. Similarly, the term ${(y-k)}^2$ can be obtained by adding the square of half the coefficient of $y$ to both sides of this equation, which yields ${(x+\frac{1}{2})}^2+(y^2+y+\frac{1}{4})=\frac{199}{2}+\frac{1}{4}+\frac{1}{4}$, or ${(x+\frac{1}{2})}^2+{(y+\frac{1}{2})}^2=\frac{199}{2}+\frac{1}{4}+\frac{1}{4}$. This equation is equivalent to ${(x+\frac{1}{2})}^2+{(y+\frac{1}{2})}^2=100$, or ${(x+\frac{1}{2})}^2+{(y+\frac{1}{2})}^2={10}^2$. Therefore, the length of the circle's radius is $10$.