Problem Analysis and Solution
The worksheet involves problems related to graphing circles. Let's solve each problem step by step.
---
####
Problem 1: Find the center and radius of the following equation:
\[
(x + 2)^2 + \left(y - \frac{1}{2}\right)^2 = 16
\]
Solution:
The standard form of the equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
Comparing the given equation:
\[
(x + 2)^2 + \left(y - \frac{1}{2}\right)^2 = 16
\]
with the standard form, we identify:
- \(h = -2\) (since \(x + 2 = x - (-2)\))
- \(k = \frac{1}{2}\)
- \(r^2 = 16\), so \(r = \sqrt{16} = 4\)
Thus, the center of the circle is \((-2, \frac{1}{2})\) and the radius is \(4\).
Correct Answer:
\[
\boxed{\left(-2, \frac{1}{2}\right), R: 4}
\]
---
####
Problem 2: Find the center and radius of the following equation:
\[
x^2 + y^2 = 30
\]
Solution:
The given equation is:
\[
x^2 + y^2 = 30
\]
This can be rewritten in the standard form of a circle's equation:
\[
(x - 0)^2 + (y - 0)^2 = 30
\]
From this, we identify:
- \(h = 0\)
- \(k = 0\)
- \(r^2 = 30\), so \(r = \sqrt{30}\)
Thus, the center of the circle is \((0, 0)\) and the radius is \(\sqrt{30}\).
Correct Answer:
\[
\boxed{(0, 0), R: \sqrt{30}}
\]
---
####
Problem 3: Choose the equation for the following circle:
The image shows a circle centered at \((-1, 3)\) with a radius of \(3\).
Solution:
The standard form of the equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
From the image:
- The center \((h, k) = (-1, 3)\)
- The radius \(r = 3\), so \(r^2 = 9\)
Substituting these values into the standard form:
\[
(x - (-1))^2 + (y - 3)^2 = 9
\]
Simplifying:
\[
(x + 1)^2 + (y - 3)^2 = 9
\]
Correct Answer:
\[
\boxed{(x + 1)^2 + (y - 3)^2 = 9}
\]
---
Final Answers:
1. \(\boxed{\left(-2, \frac{1}{2}\right), R: 4}\)
2. \(\boxed{(0, 0), R: \sqrt{30}}\)
3. \(\boxed{(x + 1)^2 + (y - 3)^2 = 9}\)
Parent Tip: Review the logic above to help your child master the concept of circle equation worksheet.
