To find the center and the radius of the circle given by the equation
[tex]\[ x^2 + y^2 - x - 2y - \frac{11}{4} = 0, \][/tex]
we need to rewrite this equation in the standard form of a circle's equation:
[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.
The equation of the circle is:
[tex]\[ x^2 + y^2 - x - 2y - \frac{11}{4} = 0. \][/tex]
First, we complete the square for the [tex]\(x\)[/tex]-terms and the [tex]\(y\)[/tex]-terms.
1. For the [tex]\(x\)[/tex]-terms:
[tex]\[ x^2 - x \][/tex]
To complete the square:
[tex]\[ x^2 - x = \left(x^2 - x + \left(\frac{1}{2}\right)^2\right) - \left(\frac{1}{2}\right)^2 = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} \][/tex]
2. For the [tex]\(y\)[/tex]-terms:
[tex]\[ y^2 - 2y \][/tex]
To complete the square:
[tex]\[ y^2 - 2y = \left(y^2 - 2y + 1\right) - 1 = \left(y - 1\right)^2 - 1 \][/tex]
Now substituting these expressions back into the original equation:
[tex]\[ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y - 1\right)^2 - 1 - \frac{11}{4} = 0 \][/tex]
Combine the constants on the right side:
[tex]\[ \left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = \frac{1}{4} + 1 + \frac{11}{4} = 4 \][/tex]
Now we have the equation in the standard form of a circle:
[tex]\[ \left(x - \frac{1}{2}\right)^2 + \left(y - 1\right)^2 = 4 \][/tex]
From this, we can identify:
- The center [tex]\((h, k)\)[/tex] of the circle is:
[tex]\[ \left( \frac{1}{2}, 1 \right) \][/tex]
- The radius [tex]\(r\)[/tex] is:
[tex]\[ \sqrt{4} = 2 \][/tex]
Therefore, the correct choice is:
C. [tex]\(\left(\frac{1}{2}, 1\right), 2\)[/tex] units
