Answer :
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 need to rewrite this equation in the standard form of a circle's equation, which is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. Here, [tex]\((h, k)\)[/tex] are the coordinates of the center, and [tex]\(r\)[/tex] is the radius.
1. Grouping the terms:
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. Completing the square for [tex]\(x\)[/tex]:
To complete the square for the [tex]\(x\)[/tex]-terms [tex]\(x^2 - x\)[/tex]:
[tex]\[ x^2 - x = (x^2 - x + \left(\frac{1}{2}\right)^2) - \left(\frac{1}{2}\right)^2 = (x - \frac{1}{2})^2 - \frac{1}{4}. \][/tex]
3. Completing the square for [tex]\(y\)[/tex]:
To complete the square for the [tex]\(y\)[/tex]-terms [tex]\(y^2 - 2y\)[/tex]:
[tex]\[ y^2 - 2y = (y^2 - 2y + 1) - 1 = (y - 1)^2 - 1. \][/tex]
4. Substitute back into the equation:
Substitute the completed squares back into the original equation:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - 1)^2 - 1 = \frac{11}{4}. \][/tex]
5. Simplify the equation:
Combine the constants on the left-hand side:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 = \frac{11}{4}. \][/tex]
Simplify the constants:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{5}{4} = \frac{11}{4}. \][/tex]
Move [tex]\(\frac{5}{4}\)[/tex] to the right-hand side:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{11}{4} + \frac{5}{4} = \frac{16}{4} = 4. \][/tex]
6. Identify the center and radius:
Now, the equation [tex]\((x - \frac{1}{2})^2 + (y - 1)^2 = 4\)[/tex] is in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\(h = \frac{1}{2}\)[/tex], [tex]\(k = 1\)[/tex], and [tex]\(r^2 = 4\)[/tex], so [tex]\(r = 2\)[/tex].
Thus, the center of the circle is [tex]\(\left(\frac{1}{2}, 1\right)\)[/tex] and the radius is [tex]\(2\)[/tex] units. Therefore, the correct option is:
D. [tex]\(\left(\frac{1}{2}, 1\right), 2\)[/tex] units.
[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, which is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. Here, [tex]\((h, k)\)[/tex] are the coordinates of the center, and [tex]\(r\)[/tex] is the radius.
1. Grouping the terms:
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. Completing the square for [tex]\(x\)[/tex]:
To complete the square for the [tex]\(x\)[/tex]-terms [tex]\(x^2 - x\)[/tex]:
[tex]\[ x^2 - x = (x^2 - x + \left(\frac{1}{2}\right)^2) - \left(\frac{1}{2}\right)^2 = (x - \frac{1}{2})^2 - \frac{1}{4}. \][/tex]
3. Completing the square for [tex]\(y\)[/tex]:
To complete the square for the [tex]\(y\)[/tex]-terms [tex]\(y^2 - 2y\)[/tex]:
[tex]\[ y^2 - 2y = (y^2 - 2y + 1) - 1 = (y - 1)^2 - 1. \][/tex]
4. Substitute back into the equation:
Substitute the completed squares back into the original equation:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - 1)^2 - 1 = \frac{11}{4}. \][/tex]
5. Simplify the equation:
Combine the constants on the left-hand side:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 = \frac{11}{4}. \][/tex]
Simplify the constants:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{5}{4} = \frac{11}{4}. \][/tex]
Move [tex]\(\frac{5}{4}\)[/tex] to the right-hand side:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{11}{4} + \frac{5}{4} = \frac{16}{4} = 4. \][/tex]
6. Identify the center and radius:
Now, the equation [tex]\((x - \frac{1}{2})^2 + (y - 1)^2 = 4\)[/tex] is in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\(h = \frac{1}{2}\)[/tex], [tex]\(k = 1\)[/tex], and [tex]\(r^2 = 4\)[/tex], so [tex]\(r = 2\)[/tex].
Thus, the center of the circle is [tex]\(\left(\frac{1}{2}, 1\right)\)[/tex] and the radius is [tex]\(2\)[/tex] units. Therefore, the correct option is:
D. [tex]\(\left(\frac{1}{2}, 1\right), 2\)[/tex] units.