Elementary Geometry For College Students, 7e
7th Edition
ISBN: 9781337614085
Author: Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher: Cengage,
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---
#### Steps and Solution:
1. Start with the given equation:
\[
x^2 + y^2 - 10x + 6y - 15 = 0
\]
2. Group the \(x\) terms and the \(y\) terms together, and move the constant term to the other side of the equation:
\[
x^2 - 10x + y^2 + 6y = 15
\]
3. Complete the square for the \(x\)-terms:
\[
x^2 - 10x \quad \text{becomes} \quad (x-5)^2 - 25 \quad \text{(since } (-5)^2 = 25 \text{)}
\]
4. Complete the square for the \(y\)-terms:
\[
y^2 + 6y \quad \text{becomes} \quad (y+3)^2 - 9 \quad \text{(since } (+3)^2 = 9 \text{)}
\]
5. Rewrite the equation with the completed squares:
\[
(x-5)^2 - 25 + (y+3)^2 - 9 = 15
\]
6. Combine the constants on the right-hand side:
\[
(x-5)^2 + (y+3)^2 - 34 = 15
\]
7. Simplify by adding 34 to both sides to balance the equation:
\[
(x-5)^2 + (y+3)^2 = 49
\]
8. The equation is now in the center-radius form \((x-h)^2 + (y-k)^2 = r^2\), where \( (h, k) \) is the center and \( r \) is the radius.
#### Identify the Center and Radius:
- **Center**: \((5, -3)\)
- **Radius**: \(7\) (since \( r^2 = 49 \), therefore](https://content.bartleby.com/qna-images/question/66ae25c8-0515-4b9c-b99a-f65507bb17d7/4d46b024-46d2-4742-8a82-c42f6965935d/honl1b_processed.jpeg)
Transcribed Image Text:### Completing the Square Method for an Equation of a Circle
#### Problem Statement:
**Complete the square method: Show all your work!**
Put \(x^2 + y^2 - 10x + 6y - 15 = 0\) in [center radius form](#).
---
#### Steps and Solution:
1. Start with the given equation:
\[
x^2 + y^2 - 10x + 6y - 15 = 0
\]
2. Group the \(x\) terms and the \(y\) terms together, and move the constant term to the other side of the equation:
\[
x^2 - 10x + y^2 + 6y = 15
\]
3. Complete the square for the \(x\)-terms:
\[
x^2 - 10x \quad \text{becomes} \quad (x-5)^2 - 25 \quad \text{(since } (-5)^2 = 25 \text{)}
\]
4. Complete the square for the \(y\)-terms:
\[
y^2 + 6y \quad \text{becomes} \quad (y+3)^2 - 9 \quad \text{(since } (+3)^2 = 9 \text{)}
\]
5. Rewrite the equation with the completed squares:
\[
(x-5)^2 - 25 + (y+3)^2 - 9 = 15
\]
6. Combine the constants on the right-hand side:
\[
(x-5)^2 + (y+3)^2 - 34 = 15
\]
7. Simplify by adding 34 to both sides to balance the equation:
\[
(x-5)^2 + (y+3)^2 = 49
\]
8. The equation is now in the center-radius form \((x-h)^2 + (y-k)^2 = r^2\), where \( (h, k) \) is the center and \( r \) is the radius.
#### Identify the Center and Radius:
- **Center**: \((5, -3)\)
- **Radius**: \(7\) (since \( r^2 = 49 \), therefore
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