Answer :
To determine the center and radius of the circle described by the equation:
[tex]\[ x^2 + y^2 - x - 2y - \frac{11}{4} = 0 \][/tex]
we will proceed by completing the square.
1. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 - x + y^2 - 2y = \frac{11}{4} \][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms:
To complete the square for [tex]\( x^2 - x \)[/tex]:
[tex]\[ x^2 - x = \left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \][/tex]
[tex]\[ = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} \][/tex]
3. Complete the square for the [tex]\(y\)[/tex] terms:
To complete the square for [tex]\( y^2 - 2y \)[/tex]:
[tex]\[ y^2 - 2y = (y - 1)^2 - 1 \][/tex]
4. Substitute these completed squares back into the equation:
[tex]\[ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + (y - 1)^2 - 1 = \frac{11}{4} \][/tex]
5. Combine constants on the right-hand side:
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 - \frac{1}{4} - 1 = \frac{11}{4} \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 - \frac{5}{4} = \frac{11}{4} \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = \frac{11}{4} + \frac{5}{4} \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = \frac{16}{4} \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = 4 \][/tex]
Now we have the equation of the circle in standard form:
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = 2^2 \][/tex]
From this equation, we can see:
- 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]\(2\)[/tex] units.
Thus, the coordinates for the center of the circle and the length of the radius are given by:
A. [tex]\(\left(\frac{1}{2}, 1\right), 2\)[/tex] units
[tex]\[ x^2 + y^2 - x - 2y - \frac{11}{4} = 0 \][/tex]
we will proceed by completing the square.
1. Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 - x + y^2 - 2y = \frac{11}{4} \][/tex]
2. Complete the square for the [tex]\(x\)[/tex] terms:
To complete the square for [tex]\( x^2 - x \)[/tex]:
[tex]\[ x^2 - x = \left(x - \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \][/tex]
[tex]\[ = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} \][/tex]
3. Complete the square for the [tex]\(y\)[/tex] terms:
To complete the square for [tex]\( y^2 - 2y \)[/tex]:
[tex]\[ y^2 - 2y = (y - 1)^2 - 1 \][/tex]
4. Substitute these completed squares back into the equation:
[tex]\[ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + (y - 1)^2 - 1 = \frac{11}{4} \][/tex]
5. Combine constants on the right-hand side:
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 - \frac{1}{4} - 1 = \frac{11}{4} \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 - \frac{5}{4} = \frac{11}{4} \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = \frac{11}{4} + \frac{5}{4} \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = \frac{16}{4} \][/tex]
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = 4 \][/tex]
Now we have the equation of the circle in standard form:
[tex]\[ \left(x - \frac{1}{2}\right)^2 + (y - 1)^2 = 2^2 \][/tex]
From this equation, we can see:
- 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]\(2\)[/tex] units.
Thus, the coordinates for the center of the circle and the length of the radius are given by:
A. [tex]\(\left(\frac{1}{2}, 1\right), 2\)[/tex] units
