Decide whether or not the equation has a circle as its graph. If it does, give the center and the radius. If it does not, describe the graph. x2 + y2 + 12x + 4y + 36 = 0 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. O A. The graph of the equation is a point. O B. The graph of the equation is a circle with center (Type an ordered pair.) The radius of the circle is OC. The graph of the equation is a line. O D. The graph is nonexistent.
Decide whether or not the equation has a circle as its graph. If it does, give the center and the radius. If it does not, describe the graph. x2 + y2 + 12x + 4y + 36 = 0 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. O A. The graph of the equation is a point. O B. The graph of the equation is a circle with center (Type an ordered pair.) The radius of the circle is OC. The graph of the equation is a line. O D. The graph is nonexistent.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Understanding Circle Equations in Coordinate Geometry
**Problem Statement:**
Decide whether or not the equation has a circle as its graph. If it does, give the center and the radius. If it does not, describe the graph.
Given equation:
\[ x^2 + y^2 + 12x + 4y + 36 = 0 \]
**Instructions:**
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
- **A.** The graph of the equation is a point.
- **B.** The graph of the equation is a circle with center \(\_\_\_\). (Type an ordered pair.)
The radius of the circle is \(\_\_\_\).
- **C.** The graph of the equation is a line.
- **D.** The graph is nonexistent.
**Detailed Explanation:**
To identify whether the given quadratic equation represents a circle, we can manipulate it into the standard form of the equation of a circle, which is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
#### Steps to Determine the Form of the Equation:
1. **Rewrite the equation**: Combine like terms involving \(x\) and \(y\).
\[ x^2 + y^2 + 12x + 4y + 36 = 0 \]
2. **Complete the square** for both \(x\) and \(y\)-terms.
For \(x\)-terms:
\[
x^2 + 12x = (x^2 + 12x + 36) - 36 = (x + 6)^2 - 36
\]
For \(y\)-terms:
\[
y^2 + 4y = (y^2 + 4y + 4) - 4 = (y + 2)^2 - 4
\]
3. **Substitute these back** into the equation and simplify:
\[
(x + 6)^2 - 36 + (y + 2)^2 - 4 + 36 = 0
\]
\[
(x + 6)^2 + (y + 2)^2 - 4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb04437b-3f7e-4496-ba03-3a7f713d70b1%2F6d508f73-5ee9-4215-9885-a5ac5a598200%2Frxzovfg_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Circle Equations in Coordinate Geometry
**Problem Statement:**
Decide whether or not the equation has a circle as its graph. If it does, give the center and the radius. If it does not, describe the graph.
Given equation:
\[ x^2 + y^2 + 12x + 4y + 36 = 0 \]
**Instructions:**
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
- **A.** The graph of the equation is a point.
- **B.** The graph of the equation is a circle with center \(\_\_\_\). (Type an ordered pair.)
The radius of the circle is \(\_\_\_\).
- **C.** The graph of the equation is a line.
- **D.** The graph is nonexistent.
**Detailed Explanation:**
To identify whether the given quadratic equation represents a circle, we can manipulate it into the standard form of the equation of a circle, which is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
#### Steps to Determine the Form of the Equation:
1. **Rewrite the equation**: Combine like terms involving \(x\) and \(y\).
\[ x^2 + y^2 + 12x + 4y + 36 = 0 \]
2. **Complete the square** for both \(x\) and \(y\)-terms.
For \(x\)-terms:
\[
x^2 + 12x = (x^2 + 12x + 36) - 36 = (x + 6)^2 - 36
\]
For \(y\)-terms:
\[
y^2 + 4y = (y^2 + 4y + 4) - 4 = (y + 2)^2 - 4
\]
3. **Substitute these back** into the equation and simplify:
\[
(x + 6)^2 - 36 + (y + 2)^2 - 4 + 36 = 0
\]
\[
(x + 6)^2 + (y + 2)^2 - 4
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