Answer :
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].
