To determine the radius of the circle represented by the equation [tex]\( x^2 + x + y^2 + y = \frac{199}{2} \)[/tex], we start by rewriting the given equation in the standard form of a circle, which is [tex]\( (x-h)^2 + (y-k)^2 = r^2 \)[/tex]. This involves completing the square for both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms.
1. Start with the given equation:
[tex]\[ x^2 + x + y^2 + y = \frac{199}{2} \][/tex]
2. To complete the square for [tex]\( x \)[/tex]:
[tex]\[
x^2 + x = \left( x + \frac{1}{2} \right)^2 - \frac{1}{4}
\][/tex]
3. To complete the square for [tex]\( y \)[/tex]:
[tex]\[
y^2 + y = \left( y + \frac{1}{2} \right)^2 - \frac{1}{4}
\][/tex]
4. Substitute these completed squares back into the original equation:
[tex]\[
\left( x + \frac{1}{2} \right)^2 - \frac{1}{4} + \left( y + \frac{1}{2} \right)^2 - \frac{1}{4} = \frac{199}{2}
\][/tex]
5. Combine constants to simplify:
[tex]\[
\left( x + \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 - \frac{1}{4} - \frac{1}{4} = \frac{199}{2}
\][/tex]
[tex]\[
\left( x + \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 - \frac{1}{2} = \frac{199}{2}
\][/tex]
6. Add [tex]\(\frac{1}{2}\)[/tex] to both sides of the equation to balance it:
[tex]\[
\left( x + \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 = \frac{199}{2} + \frac{1}{2}
\][/tex]
[tex]\[
\left( x + \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 = \frac{200}{2}
\][/tex]
[tex]\[
\left( x + \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 = 100
\][/tex]
7. The equation is now in the standard form [tex]\( (x-h)^2 + (y-k)^2 = r^2 \)[/tex], where [tex]\( h = -\frac{1}{2} \)[/tex], [tex]\( k = -\frac{1}{2} \)[/tex], and [tex]\( r^2 = 100 \)[/tex].
8. To find the radius [tex]\( r \)[/tex], take the square root of [tex]\( r^2 \)[/tex]:
[tex]\[
r = \sqrt{100} = 10
\][/tex]
Thus, the radius of the circle is [tex]\( \boxed{10} \)[/tex].
