Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Ask your questions and receive precise answers from experienced professionals across different disciplines. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.