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To determine the center and radius of the circle given by the equation
[tex]\[ x^2 + y^2 - x - 2y - \frac{11}{4} = 0, \][/tex]
we need to write this equation in the standard form of a circle's equation. The standard form 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.
Step 1: Group the [tex]\(x\)[/tex] terms and [tex]\(y\)[/tex] terms together:
[tex]\[ (x^2 - x) + (y^2 - 2y) = \frac{11}{4}. \][/tex]
Step 2: Complete the square for the [tex]\(x\)[/tex] terms:
- Take the coefficient of [tex]\(x\)[/tex] which is [tex]\(-1\)[/tex], divide it by 2, giving [tex]\(-\frac{1}{2}\)[/tex], and then square it, obtaining [tex]\(\left(-\frac{1}{2}\right)^2 = \frac{1}{4}\)[/tex].
Add and subtract [tex]\(\frac{1}{4}\)[/tex] inside the bracket:
[tex]\[ x^2 - x + \frac{1}{4} - \frac{1}{4}. \][/tex]
Rewrite as a perfect square trinomial:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4}. \][/tex]
Step 3: Complete the square for the [tex]\(y\)[/tex] terms:
- Take the coefficient of [tex]\(y\)[/tex] which is [tex]\(-2\)[/tex], divide it by 2, giving [tex]\(-1\)[/tex], and then square it, obtaining [tex]\(\left(-1\right)^2 = 1\)[/tex].
Add and subtract 1 inside the bracket:
[tex]\[ y^2 - 2y + 1 - 1. \][/tex]
Rewrite as a perfect square trinomial:
[tex]\[ (y - 1)^2 - 1. \][/tex]
Step 4: Rewrite the entire equation with the completed squares:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - 1)^2 - 1 = \frac{11}{4}. \][/tex]
Combine constants on the right side:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 = \frac{11}{4}. \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{11}{4} + \frac{1}{4} + 1. \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{11}{4} + \frac{1}{4} + \frac{4}{4} = \frac{16}{4} = 4. \][/tex]
So, we now have the equation of the circle in standard form:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{13}{4}. \][/tex]
From this, we can deduce 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]\(\sqrt{\frac{13}{4}}\)[/tex].
Calculating the radius:
[tex]\[ r = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \approx 1.8027756377319946. \][/tex]
Therefore, the correct center and radius of the circle are [tex]\(\left(\frac{1}{2}, 1\right)\)[/tex] and [tex]\(\frac{\sqrt{13}}{2} \approx 1.8027756377319946\)[/tex].
Thus, none of the provided options exactly match the radius, but the correct one should be the one with the right center. Therefore, the correct option is:
A. [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 write this equation in the standard form of a circle's equation. The standard form 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.
Step 1: Group the [tex]\(x\)[/tex] terms and [tex]\(y\)[/tex] terms together:
[tex]\[ (x^2 - x) + (y^2 - 2y) = \frac{11}{4}. \][/tex]
Step 2: Complete the square for the [tex]\(x\)[/tex] terms:
- Take the coefficient of [tex]\(x\)[/tex] which is [tex]\(-1\)[/tex], divide it by 2, giving [tex]\(-\frac{1}{2}\)[/tex], and then square it, obtaining [tex]\(\left(-\frac{1}{2}\right)^2 = \frac{1}{4}\)[/tex].
Add and subtract [tex]\(\frac{1}{4}\)[/tex] inside the bracket:
[tex]\[ x^2 - x + \frac{1}{4} - \frac{1}{4}. \][/tex]
Rewrite as a perfect square trinomial:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4}. \][/tex]
Step 3: Complete the square for the [tex]\(y\)[/tex] terms:
- Take the coefficient of [tex]\(y\)[/tex] which is [tex]\(-2\)[/tex], divide it by 2, giving [tex]\(-1\)[/tex], and then square it, obtaining [tex]\(\left(-1\right)^2 = 1\)[/tex].
Add and subtract 1 inside the bracket:
[tex]\[ y^2 - 2y + 1 - 1. \][/tex]
Rewrite as a perfect square trinomial:
[tex]\[ (y - 1)^2 - 1. \][/tex]
Step 4: Rewrite the entire equation with the completed squares:
[tex]\[ (x - \frac{1}{2})^2 - \frac{1}{4} + (y - 1)^2 - 1 = \frac{11}{4}. \][/tex]
Combine constants on the right side:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 - \frac{1}{4} - 1 = \frac{11}{4}. \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{11}{4} + \frac{1}{4} + 1. \][/tex]
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{11}{4} + \frac{1}{4} + \frac{4}{4} = \frac{16}{4} = 4. \][/tex]
So, we now have the equation of the circle in standard form:
[tex]\[ (x - \frac{1}{2})^2 + (y - 1)^2 = \frac{13}{4}. \][/tex]
From this, we can deduce 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]\(\sqrt{\frac{13}{4}}\)[/tex].
Calculating the radius:
[tex]\[ r = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \approx 1.8027756377319946. \][/tex]
Therefore, the correct center and radius of the circle are [tex]\(\left(\frac{1}{2}, 1\right)\)[/tex] and [tex]\(\frac{\sqrt{13}}{2} \approx 1.8027756377319946\)[/tex].
Thus, none of the provided options exactly match the radius, but the correct one should be the one with the right center. Therefore, the correct option is:
A. [tex]\(\left(\frac{1}{2}, 1\right), 2\)[/tex] units
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