To find the coordinates for the center of the circle and the length of the radius for the equation [tex]\( x^2 + y^2 - x - 2y - \frac{11}{4} = 0 \)[/tex], we will follow these steps:
1. Rewrite the equation in the form [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex]:
Start by grouping the [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms separately:
[tex]\[
x^2 - x + y^2 - 2y = \frac{11}{4}
\][/tex]
2. Complete the square for the [tex]\(x\)[/tex]-terms:
- For [tex]\(x^2 - x\)[/tex], take half the coefficient of [tex]\(x\)[/tex] (which is [tex]\(-1\)[/tex]), square it, and add and subtract that value.
[tex]\[
x^2 - x \quad \text{becomes} \quad (x - \frac{1}{2})^2 - \left(\frac{1}{2}\right)^2
\][/tex]
3. Complete the square for the [tex]\(y\)[/tex]-terms:
- For [tex]\(y^2 - 2y\)[/tex], take half the coefficient of [tex]\(y\)[/tex] (which is [tex]\(-2\)[/tex]), square it, and add and subtract that value.
[tex]\[
y^2 - 2y \quad \text{becomes} \quad (y - 1)^2 - 1^2
\][/tex]
4. Substitute the completed squares into the equation:
[tex]\[
(x - \frac{1}{2})^2 - \left(\frac{1}{2}\right)^2 + (y - 1)^2 - 1 = \frac{11}{4}
\][/tex]
Simplify the constants on the right-hand side:
[tex]\[
(x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 = \frac{11}{4}
\][/tex]
Combine the constants:
[tex]\[
(x - \frac{1}{2})^2 + (y - 1)^2 - \frac{5}{4} = \frac{11}{4}
\][/tex]
Add [tex]\(\frac{5}{4}\)[/tex] to both sides:
[tex]\[
(x - \frac{1}{2})^2 + (y - 1)^2 = \frac{11}{4} + \frac{5}{4} = 4
\][/tex]
5. Identify the center and radius:
- The equation [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex] reveals that the center is [tex]\((h, k)\)[/tex] and the radius is [tex]\(r\)[/tex].
- Here, [tex]\((x - \frac{1}{2})^2 + (y - 1)^2 = 4\)[/tex] implies the center [tex]\(\left( \frac{1}{2}, 1 \right)\)[/tex] and the radius is [tex]\(\sqrt{4} = 2\)[/tex].
Thus, the coordinates for the center of the circle and the length of the radius are:
Center: [tex]\(\left( \frac{1}{2}, 1 \right)\)[/tex]
Radius: [tex]\(2\)[/tex] units
So, the correct answer is:
A. [tex]\(\left( \frac{1}{2}, 1 \right), 2\)[/tex] units
