To find the coordinates for the center of the circle and the length of the radius, let's convert the given equation [tex]\( x^2 + y^2 - x - 2y - \frac{11}{4} = 0 \)[/tex] into the standard form of a circle's equation.
The standard form of a circle's equation is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
Given equation:
[tex]\[
x^2 + y^2 - x - 2y - \frac{11}{4} = 0
\][/tex]
We start by grouping the [tex]\(x\)[/tex] terms and the [tex]\(y\)[/tex] terms together and moving the constant term to the other side of the equation:
[tex]\[
x^2 - x + y^2 - 2y = \frac{11}{4}
\][/tex]
Next, we complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
For [tex]\(x^2 - x\)[/tex]:
1. Take half of the coefficient of [tex]\(x\)[/tex] (which is [tex]\(-1\)[/tex]), square it, and add it inside the equation:
[tex]\[
x^2 - x = \left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4}
\][/tex]
For [tex]\(y^2 - 2y\)[/tex]:
1. Take half of the coefficient of [tex]\(y\)[/tex] (which is [tex]\(-2\)[/tex]), square it, and add it inside the equation:
[tex]\[
y^2 - 2y = \left(y - 1\right)^2 - 1
\][/tex]
Now, substituting the completed squares back in:
[tex]\[
\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y - 1\right)^2 - 1 = \frac{11}{4}
\][/tex]
Combine and simplify the equation:
[tex]\[
\left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 - \frac{5}{4} = \frac{11}{4}
\][/tex]
[tex]\[
\left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = \frac{16}{4} = 4
\][/tex]
Therefore, the equation in the standard form is:
[tex]\[
\left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = 4
\][/tex]
From this, we can determine that:
- The center of the circle [tex]\((h, k)\)[/tex] is [tex]\(\left(\frac{1}{2}, 1\right)\)[/tex]
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{4} = 2\)[/tex] units
Thus, the correct answer is:
D. [tex]\(\left(\frac{1}{2}, 1\right), 2\)[/tex] units
