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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Question
![Below is a transcription of the question for educational purposes:
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**Question 11:**
What are the coordinates of the center and the length of the radius of the circle represented by the equation \( x^2 + y^2 - 4x + 8y + 11 = 0 \)?
Options:
1. Center \((2, -4)\) and radius 3
2. Center \((-2, 4)\) and radius 3
3. Center \((2, -4)\) and radius 9
4. Center \((-2, 4)\) and radius 9
---
To solve this problem, one needs to rewrite the given equation of the circle in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. This involves completing the square for both \(x\) and \(y\).
Steps to complete the square:
1. Group the \(x\) and \(y\) terms: \(x^2 - 4x + y^2 + 8y + 11 = 0\).
2. Complete the square for \(x\): \(x^2 - 4x \rightarrow (x - 2)^2 - 4\).
3. Complete the square for \(y\): \(y^2 + 8y \rightarrow (y + 4)^2 - 16\).
4. Rewrite the equation, including the constants: \((x - 2)^2 - 4 + (y + 4)^2 - 16 + 11 = 0\), which simplifies to \((x - 2)^2 + (y + 4)^2 - 9 = 0\).
The simplified equation becomes:
\[
(x - 2)^2 + (y + 4)^2 = 9
\]
Here, the center of the circle is \((2, -4)\) and the radius is \(\sqrt{9} = 3\).
Therefore, the correct answer is:
1. **Center (2, -4) and radius 3**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42779bdc-66e8-47e8-8a59-0cd69b39b8c3%2F2be598de-f77e-4025-a3d2-a24f14f78897%2Fdcwrj9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Below is a transcription of the question for educational purposes:
---
**Question 11:**
What are the coordinates of the center and the length of the radius of the circle represented by the equation \( x^2 + y^2 - 4x + 8y + 11 = 0 \)?
Options:
1. Center \((2, -4)\) and radius 3
2. Center \((-2, 4)\) and radius 3
3. Center \((2, -4)\) and radius 9
4. Center \((-2, 4)\) and radius 9
---
To solve this problem, one needs to rewrite the given equation of the circle in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. This involves completing the square for both \(x\) and \(y\).
Steps to complete the square:
1. Group the \(x\) and \(y\) terms: \(x^2 - 4x + y^2 + 8y + 11 = 0\).
2. Complete the square for \(x\): \(x^2 - 4x \rightarrow (x - 2)^2 - 4\).
3. Complete the square for \(y\): \(y^2 + 8y \rightarrow (y + 4)^2 - 16\).
4. Rewrite the equation, including the constants: \((x - 2)^2 - 4 + (y + 4)^2 - 16 + 11 = 0\), which simplifies to \((x - 2)^2 + (y + 4)^2 - 9 = 0\).
The simplified equation becomes:
\[
(x - 2)^2 + (y + 4)^2 = 9
\]
Here, the center of the circle is \((2, -4)\) and the radius is \(\sqrt{9} = 3\).
Therefore, the correct answer is:
1. **Center (2, -4) and radius 3**
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