Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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help plz i need it and thanks
![**Determine the Equation of the Circle Graphed Below**
---
In the graph displayed, a circle is plotted on the Cartesian plane. The coordinates are labeled, and axes are marked as \( x \) and \( y \). The circle's center is denoted at the coordinate point \((4, 1)\) and a point on the circumference of the circle is marked at \((2, -4)\).
### Steps to Determine the Equation of the Circle
To find the equation of the circle, we use the standard form of the equation of a circle:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Here, \((h, k)\) represents the coordinates of the circle's center, and \(r\) is the radius of the circle.
1. **Identify the Center:**
The center of the circle is given as \((4, 1)\). Therefore, \(h = 4\) and \(k = 1\).
2. **Calculate the Radius:**
The radius \(r\) is the distance between the center \((4, 1)\) and a point on the circle \((2, -4)\).
Using the distance formula:
\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
where \((x_1, y_1) = (4, 1)\) and \((x_2, y_2) = (2, -4)\).
Substituting the values:
\[
r = \sqrt{(2 - 4)^2 + (-4 - 1)^2}
= \sqrt{(-2)^2 + (-5)^2}
= \sqrt{4 + 25}
= \sqrt{29}
\]
Thus, \(r = \sqrt{29}\).
3. **Form the Equation:**
Substituting \(h = 4\), \(k = 1\), and \(r^2 = 29\) into the standard equation:
\[
(x - 4)^2 + (y - 1)^2 = 29
\]
Therefore, the equation of the circle graphed is:
\[
(x - 4)^2 + (y -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7d50051-e71b-42fe-8b2e-24c242a2c52b%2F783e7163-d505-457f-8eb8-73befb5465f0%2Flwfo92_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Determine the Equation of the Circle Graphed Below**
---
In the graph displayed, a circle is plotted on the Cartesian plane. The coordinates are labeled, and axes are marked as \( x \) and \( y \). The circle's center is denoted at the coordinate point \((4, 1)\) and a point on the circumference of the circle is marked at \((2, -4)\).
### Steps to Determine the Equation of the Circle
To find the equation of the circle, we use the standard form of the equation of a circle:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Here, \((h, k)\) represents the coordinates of the circle's center, and \(r\) is the radius of the circle.
1. **Identify the Center:**
The center of the circle is given as \((4, 1)\). Therefore, \(h = 4\) and \(k = 1\).
2. **Calculate the Radius:**
The radius \(r\) is the distance between the center \((4, 1)\) and a point on the circle \((2, -4)\).
Using the distance formula:
\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
where \((x_1, y_1) = (4, 1)\) and \((x_2, y_2) = (2, -4)\).
Substituting the values:
\[
r = \sqrt{(2 - 4)^2 + (-4 - 1)^2}
= \sqrt{(-2)^2 + (-5)^2}
= \sqrt{4 + 25}
= \sqrt{29}
\]
Thus, \(r = \sqrt{29}\).
3. **Form the Equation:**
Substituting \(h = 4\), \(k = 1\), and \(r^2 = 29\) into the standard equation:
\[
(x - 4)^2 + (y - 1)^2 = 29
\]
Therefore, the equation of the circle graphed is:
\[
(x - 4)^2 + (y -
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