Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.