Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
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.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.