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Complete the square method: Show all your work! Put x² + y² 10x + 6y 15 = 0 in center radius form. Identify its center:_ Ans.(x - 5)² + (y + 3)² = 49 c=(5, -3) r=7 radius length:

Complete the square method: Show all your work! Put x² + y² 10x + 6y 15 = 0 in center radius form. Identify its center:_ Ans.(x - 5)² + (y + 3)² = 49 c=(5, -3) r=7 radius length:

Elementary Geometry For College Students, 7e
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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### 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
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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