answer ### Identifying the Center and Radius of Each EquationWhen given an equation of the form $(x - h)^2 + (y - k)^2 = r^2$, you can determine the center and radius of the circle:- The center of the circle is $(h, k)$.- The radius of the circle is $r$, which is the square root of $r^2$.**Note**: - If $r^2$ is not a perfect square, leave $r$ in its simplified radical form, but use the decimal equivalent for graphing. - Example: $\sqrt{12} = 2\sqrt{3} = 3.46$.Here are the equations along with their corresponding centers and radii.| Eq. # | Standard Graphing Equation | Radius | Center ||-------|-------------------------------------------|--------|----------|| 1 | $(x - 2)^2 + (y - 5)^2 = 144$ | 12 | (2, 5) || 2 | $(x - 0)^2 + (y - 7)^2 = 625$ | 25 | (0, 7) || 3 | $(x - 15)^2 + (y + 8)^2 = 289$ | 17 | (15, -8) || 4 | $(x + 11)^2 + (y + 6)^2 = 324$ | 18 | (-11, -6)|| 5 | $(x + 4)^2 + (y - 9)^2 = 196$ | 14 | (-4, 9) |This table helps summarize and identify the important attributes of each circle based on the given standard form equations.

A circle is a two dimensional shape and it is closed curve. A set of all the point are at the same…

A) Identify the center and radius of each equation below. Note: Ifr² is not a perfect square then leave r in simplified radical form but use the decimal equivalent for graphing. Example: J12 = 2/3 = 3.46 Standard Graphing Equation |(x-2 )* + (y- 5 )° = 144 (x-0)²+ (y-7 )’ = 625 |(x-15)² + (y+8 )² = 289 (x+ 11)’ + (y + 6 )² = 324 |(x+4)°+ (y-9 )’ = 196 Eq. # Radius Center 1 2 %3D 5 3. 4.

A) Identify the center and radius of each equation below. Note: Ifr² is not a perfect square then leave r in simplified radical form but use the decimal equivalent for graphing. Example: J12 = 2/3 = 3.46 Standard Graphing Equation |(x-2 )* + (y- 5 )° = 144 (x-0)²+ (y-7 )’ = 625 |(x-15)² + (y+8 )² = 289 (x+ 11)’ + (y + 6 )² = 324 |(x+4)°+ (y-9 )’ = 196 Eq. # Radius Center 1 2 %3D 5 3. 4.

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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### Identifying the Center and Radius of Each Equation

When given an equation of the form $(x - h)^2 + (y - k)^2 = r^2$, you can determine the center and radius of the circle:
- The center of the circle is $(h, k)$.
- The radius of the circle is $r$, which is the square root of $r^2$.

**Note**:
- If $r^2$ is not a perfect square, leave $r$ in its simplified radical form, but use the decimal equivalent for graphing.
- Example: $\sqrt{12} = 2\sqrt{3} = 3.46$.

Here are the equations along with their corresponding centers and radii.

| Eq. # | Standard Graphing Equation | Radius | Center |
|-------|-------------------------------------------|--------|----------|
| 1 | $(x - 2)^2 + (y - 5)^2 = 144$ | 12 | (2, 5) |
| 2 | $(x - 0)^2 + (y - 7)^2 = 625$ | 25 | (0, 7) |
| 3 | $(x - 15)^2 + (y + 8)^2 = 289$ | 17 | (15, -8) |
| 4 | $(x + 11)^2 + (y + 6)^2 = 324$ | 18 | (-11, -6)|
| 5 | $(x + 4)^2 + (y - 9)^2 = 196$ | 14 | (-4, 9) |

This table helps summarize and identify the important attributes of each circle based on the given standard form equations.

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