Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To solve for the points that lie on the given circle, we start by understanding the given equation of the circle:
[tex]\[ (x - 1)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
This equation represents a circle in the standard form [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.
1. Identify the center [tex]\((h,k)\)[/tex] and radius [tex]\(r\)[/tex] of the circle:
By comparing the given circle equation to the standard form:
[tex]\[ (x - 1)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
We find:
- [tex]\(h = 1\)[/tex]
- [tex]\(k = -\frac{1}{2}\)[/tex]
The radius [tex]\(r\)[/tex] is found by taking the square root of the right-hand side:
[tex]\[ r = \sqrt{\frac{36}{25}} = \frac{6}{5} \][/tex]
2. Choose specific [tex]\( x \)[/tex] values and solve for corresponding [tex]\( y \)[/tex] values:
Let's find points lying on the circle. For simplicity, we choose [tex]\( x \)[/tex] values that are easy to work with. Let's use [tex]\( x = 1+\frac{6}{5} \)[/tex] and [tex]\( x = 1-\frac{6}{5} \)[/tex], which are at the radius distance away from the center along the x-direction.
- For [tex]\( x = 1 + \frac{6}{5} = \frac{11}{5} \)[/tex]:
Substitute [tex]\( x = \frac{11}{5} \)[/tex] into the circle's equation and solve for [tex]\( y \)[/tex]:
[tex]\[ \left(\frac{11}{5} - 1\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(\frac{11}{5} - \frac{5}{5}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(\frac{6}{5}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \frac{36}{25} + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(y + \frac{1}{2}\right)^2 = 0 \][/tex]
[tex]\[ y + \frac{1}{2} = 0 \quad \text{or} \quad y + \frac{1}{2} = 0 \][/tex]
[tex]\[ y = -\frac{1}{2} \][/tex]
So one point is [tex]\( \left(\frac{11}{5}, -\frac{1}{2}\right) \)[/tex].
- For [tex]\( x = 1 - \frac{6}{5} = -\frac{1}{5} \)[/tex]:
Substitute [tex]\( x = -\frac{1}{5} \)[/tex]:
[tex]\[ \left(-\frac{1}{5} - 1\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(-\frac{1}{5} - \frac{5}{5}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(-\frac{6}{5}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \frac{36}{25} + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(y + \frac{1}{2}\right)^2 = 0 \][/tex]
[tex]\[ y = -\frac{1}{2} \][/tex]
So another point is [tex]\( \left(-\frac{1}{5}, -\frac{1}{2}\right) \)[/tex].
3. Conclusion:
We have calculated the points:
- [tex]\( \left(\frac{11}{5}, -\frac{1}{2}\right) \)[/tex]
- [tex]\( \left(-\frac{1}{5}, -\frac{1}{2}\right) \)[/tex]
These points lie on the circle defined by the given equation [tex]\( (x - 1)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \)[/tex].
[tex]\[ (x - 1)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
This equation represents a circle in the standard form [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.
1. Identify the center [tex]\((h,k)\)[/tex] and radius [tex]\(r\)[/tex] of the circle:
By comparing the given circle equation to the standard form:
[tex]\[ (x - 1)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
We find:
- [tex]\(h = 1\)[/tex]
- [tex]\(k = -\frac{1}{2}\)[/tex]
The radius [tex]\(r\)[/tex] is found by taking the square root of the right-hand side:
[tex]\[ r = \sqrt{\frac{36}{25}} = \frac{6}{5} \][/tex]
2. Choose specific [tex]\( x \)[/tex] values and solve for corresponding [tex]\( y \)[/tex] values:
Let's find points lying on the circle. For simplicity, we choose [tex]\( x \)[/tex] values that are easy to work with. Let's use [tex]\( x = 1+\frac{6}{5} \)[/tex] and [tex]\( x = 1-\frac{6}{5} \)[/tex], which are at the radius distance away from the center along the x-direction.
- For [tex]\( x = 1 + \frac{6}{5} = \frac{11}{5} \)[/tex]:
Substitute [tex]\( x = \frac{11}{5} \)[/tex] into the circle's equation and solve for [tex]\( y \)[/tex]:
[tex]\[ \left(\frac{11}{5} - 1\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(\frac{11}{5} - \frac{5}{5}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(\frac{6}{5}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \frac{36}{25} + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(y + \frac{1}{2}\right)^2 = 0 \][/tex]
[tex]\[ y + \frac{1}{2} = 0 \quad \text{or} \quad y + \frac{1}{2} = 0 \][/tex]
[tex]\[ y = -\frac{1}{2} \][/tex]
So one point is [tex]\( \left(\frac{11}{5}, -\frac{1}{2}\right) \)[/tex].
- For [tex]\( x = 1 - \frac{6}{5} = -\frac{1}{5} \)[/tex]:
Substitute [tex]\( x = -\frac{1}{5} \)[/tex]:
[tex]\[ \left(-\frac{1}{5} - 1\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(-\frac{1}{5} - \frac{5}{5}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(-\frac{6}{5}\right)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \frac{36}{25} + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \][/tex]
[tex]\[ \left(y + \frac{1}{2}\right)^2 = 0 \][/tex]
[tex]\[ y = -\frac{1}{2} \][/tex]
So another point is [tex]\( \left(-\frac{1}{5}, -\frac{1}{2}\right) \)[/tex].
3. Conclusion:
We have calculated the points:
- [tex]\( \left(\frac{11}{5}, -\frac{1}{2}\right) \)[/tex]
- [tex]\( \left(-\frac{1}{5}, -\frac{1}{2}\right) \)[/tex]
These points lie on the circle defined by the given equation [tex]\( (x - 1)^2 + \left(y + \frac{1}{2}\right)^2 = \frac{36}{25} \)[/tex].
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. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
