Which equation represents the circle described? The radius is 2 units. The center is the same as the center of a circle whose equation is x² + y² - 8x - 6y + 24 = 0. (x + 4)² + (y + 3)² = 2 (x-4)² + (y - 3)² = 2 O (x-4)² + (y - 3)² = 2² O (x+4)² + (y + 3)² = 2²
Which equation represents the circle described? The radius is 2 units. The center is the same as the center of a circle whose equation is x² + y² - 8x - 6y + 24 = 0. (x + 4)² + (y + 3)² = 2 (x-4)² + (y - 3)² = 2 O (x-4)² + (y - 3)² = 2² O (x+4)² + (y + 3)² = 2²
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
![**Question: Which equation represents the circle described?**
- The radius is 2 units.
- The center is the same as the center of a circle whose equation is \( x^2 + y^2 - 8x - 6y + 24 = 0 \).
**Options:**
- \( (x + 4)^2 + (y + 3)^2 = 2 \)
- \( (x - 4)^2 + (y - 3)^2 = 2 \) ⬅️ (Selected)
- \( (x - 4)^2 + (y - 3)^2 = 2^2 \)
- \( (x + 4)^2 + (y + 3)^2 = 2^2 \)
In the above question, students are asked to determine the correct equation of a circle based on the given radius and center. The provided details help in identifying the correct equation.
### Explanation:
1. **Determine the center of the given circle equation:**
The given circle's equation is \( x^2 + y^2 - 8x - 6y + 24 = 0 \). We need to rewrite this in the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), where (h, k) is the center and r is the radius.
- Complete the square for \( x \) and \( y \):
\[
x^2 - 8x \implies (x - 4)^2 - 16
\]
\[
y^2 - 6y \implies (y - 3)^2 - 9
\]
So, the given equation becomes:
\[
(x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0
\]
Simplify it:
\[
(x - 4)^2 + (y - 3)^2 - 1 = 0 \implies (x - 4)^2 + (y - 3)^2 = 1
\]
Therefore, the center of the existing circle is \((4, 3)\), and its radius is \(\sqrt{1} = 1\).
2. **Determine the new circle's equation:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F97313d02-0661-4466-b253-1f053257a28a%2F87ddc2b2-1a02-450d-89c0-0c784a1ae3e9%2F3bym0qf_processed.png&w=3840&q=75)
Transcribed Image Text:**Question: Which equation represents the circle described?**
- The radius is 2 units.
- The center is the same as the center of a circle whose equation is \( x^2 + y^2 - 8x - 6y + 24 = 0 \).
**Options:**
- \( (x + 4)^2 + (y + 3)^2 = 2 \)
- \( (x - 4)^2 + (y - 3)^2 = 2 \) ⬅️ (Selected)
- \( (x - 4)^2 + (y - 3)^2 = 2^2 \)
- \( (x + 4)^2 + (y + 3)^2 = 2^2 \)
In the above question, students are asked to determine the correct equation of a circle based on the given radius and center. The provided details help in identifying the correct equation.
### Explanation:
1. **Determine the center of the given circle equation:**
The given circle's equation is \( x^2 + y^2 - 8x - 6y + 24 = 0 \). We need to rewrite this in the standard form \( (x - h)^2 + (y - k)^2 = r^2 \), where (h, k) is the center and r is the radius.
- Complete the square for \( x \) and \( y \):
\[
x^2 - 8x \implies (x - 4)^2 - 16
\]
\[
y^2 - 6y \implies (y - 3)^2 - 9
\]
So, the given equation becomes:
\[
(x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0
\]
Simplify it:
\[
(x - 4)^2 + (y - 3)^2 - 1 = 0 \implies (x - 4)^2 + (y - 3)^2 = 1
\]
Therefore, the center of the existing circle is \((4, 3)\), and its radius is \(\sqrt{1} = 1\).
2. **Determine the new circle's equation:**
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