To find the coordinates of the center of the circle and the radius, we start by rewriting the given circle equation in its standard form. The original equation is:
[tex]\[ x^2 + y^2 - x - 2y - \frac{11}{4} = 0 \][/tex]
First, let's complete the square for both [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
1. Rearrange the equation so that all [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms are on the left side:
[tex]\[ x^2 - x + y^2 - 2y = \frac{11}{4} \][/tex]
2. Complete the square for the [tex]\( x \)[/tex]-terms and the [tex]\( y \)[/tex]-terms separately:
- For the [tex]\( x \)[/tex]-terms: [tex]\( x^2 - x \)[/tex]:
[tex]\[ x^2 - x = \left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \][/tex]
- For the [tex]\( y \)[/tex]-terms: [tex]\( y^2 - 2y \)[/tex]:
[tex]\[ y^2 - 2y = \left(y - 1\right)^2 - 1 \][/tex]
3. Substitute these completed squares back into the equation:
[tex]\[ \left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + \left(y - 1\right)^2 - 1 = \frac{11}{4} \][/tex]
4. Simplify the equation by combining constants on the right side:
[tex]\[ \left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = \frac{11}{4} + \frac{1}{4} + 1 \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = 4 \][/tex]
Now, the equation is in the standard form of a circle [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius.
From the equation:
[tex]\[ \left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = 4 \][/tex]
- The center [tex]\((h, k)\)[/tex] of the circle is: [tex]\( \left( \frac{1}{2}, 1 \right) \)[/tex]
- The radius [tex]\( r \)[/tex] of the circle is: [tex]\( \sqrt{4} = 2 \)[/tex]
The coordinates for the center of the circle and the length of the radius are:
[tex]\[ \boxed{\left( \frac{1}{2}, 1 \right), 2 \text{ units}} \][/tex]
So the correct answer is:
B. [tex]\(\left(\frac{1}{2}, 1\right), 2\)[/tex] units
