24 What are the center and radius of the circle 2 +y² + 4x = 5 ? whose equation is 1) (2,0) and 1 2) (-2,0) and 1 3) (2,0) and 3 4) (-2,0) and 3
24 What are the center and radius of the circle 2 +y² + 4x = 5 ? whose equation is 1) (2,0) and 1 2) (-2,0) and 1 3) (2,0) and 3 4) (-2,0) and 3
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 24:** What are the center and radius of the circle whose equation is \( x^2 + y^2 + 4x = 5 \)?
1) \( (2,0) \) and 1
2) \( (-2,0) \) and 1
3) \( (2,0) \) and 3
4) \( (-2,0) \) and 3
**Explanation:**
To determine the center and radius of the circle given by the equation \( x^2 + y^2 + 4x = 5 \), we need to rewrite the equation in the standard form of a circle.
1. Start with the equation:
\[ x^2 + y^2 + 4x = 5 \]
2. Complete the square for the \( x \) terms:
\[ x^2 + 4x + y^2 = 5 \]
3. Add and subtract the square of half the coefficient of \( x \) (which is 2):
\[ x^2 + 4x + 4 + y^2 = 5 + 4 \]
\[ (x + 2)^2 + y^2 = 9 \]
Now the equation is in standard form \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) is the center and \( r \) is the radius.
From \( (x + 2)^2 + y^2 = 9 \), we can see that:
- The center \((h, k) = (-2, 0)\)
- The radius \( r = \sqrt{9} = 3 \)
Thus, the correct answer is:
4) \( (-2,0) \) and 3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42779bdc-66e8-47e8-8a59-0cd69b39b8c3%2F1ac6a24c-c4a6-4330-bb4c-bd5b43b01367%2Fevw23ln_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 24:** What are the center and radius of the circle whose equation is \( x^2 + y^2 + 4x = 5 \)?
1) \( (2,0) \) and 1
2) \( (-2,0) \) and 1
3) \( (2,0) \) and 3
4) \( (-2,0) \) and 3
**Explanation:**
To determine the center and radius of the circle given by the equation \( x^2 + y^2 + 4x = 5 \), we need to rewrite the equation in the standard form of a circle.
1. Start with the equation:
\[ x^2 + y^2 + 4x = 5 \]
2. Complete the square for the \( x \) terms:
\[ x^2 + 4x + y^2 = 5 \]
3. Add and subtract the square of half the coefficient of \( x \) (which is 2):
\[ x^2 + 4x + 4 + y^2 = 5 + 4 \]
\[ (x + 2)^2 + y^2 = 9 \]
Now the equation is in standard form \( (x-h)^2 + (y-k)^2 = r^2 \), where \((h, k)\) is the center and \( r \) is the radius.
From \( (x + 2)^2 + y^2 = 9 \), we can see that:
- The center \((h, k) = (-2, 0)\)
- The radius \( r = \sqrt{9} = 3 \)
Thus, the correct answer is:
4) \( (-2,0) \) and 3
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