To find the coordinates of 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 aim to rewrite it in the standard form of the equation of a circle, which is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
This requires us to complete the square for the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms.
1. Starting from the given equation:
[tex]\[ x^2 + y^2 - x - 2y - \frac{11}{4} = 0 \][/tex]
2. Focus on the [tex]\( x \)[/tex]-terms first.
[tex]\[ x^2 - x \][/tex]
To complete the square, we take half the coefficient of [tex]\( x \)[/tex] (which is -1), square it, and add and subtract this value:
[tex]\[ x^2 - x + \left( \frac{-1}{2} \right)^2 - \left( \frac{-1}{2} \right)^2 \][/tex]
[tex]\[ = (x - \frac{1}{2})^2 - \frac{1}{4} \][/tex]
3. Now, do the same for the [tex]\( y \)[/tex]-terms:
[tex]\[ y^2 - 2y \][/tex]
Take half the coefficient of [tex]\( y \)[/tex] (which is -2), square it, and add and subtract this value:
[tex]\[ y^2 - 2y + \left( \frac{-2}{2} \right)^2 - \left( \frac{-2}{2} \right)^2 \][/tex]
[tex]\[ = (y - 1)^2 - 1 \][/tex]
4. Substitute these completed squares back into the equation:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - 1)^2 - 1 - \frac{11}{4} = 0 \][/tex]
5. Combine the constants on the left-hand side of the equation:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 - \frac{11}{4} = 0 \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{1}{4} + 1 + \frac{11}{4} \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = 1 + \frac{11}{4} \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = 1 + 2.75 \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = 4 \][/tex]
6. Simplify:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = 2^2 \][/tex]
From the standard form of the circle equation [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we can see that:
- The center [tex]\((h, k)\)[/tex] is [tex]\((\frac{1}{2}, 1)\)[/tex].
- The radius [tex]\(r\)[/tex] is [tex]\(2\)[/tex].
Thus, the coordinates of the center of the circle are [tex]\(\left( \frac{1}{2}, 1 \right)\)[/tex] and the radius is [tex]\(2\)[/tex] units.
Therefore, the correct answer is:
B. [tex]\(\left( \frac{1}{2}, 1 \right), 2\)[/tex] units
