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.
"
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 line.
C. The graph of the equation is a circle with center
D. The radius of the circle is
D. The graph is nonexistent.

Ask by Sanders Ball. in the United States

Jan 21,2025

Question

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.
"
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 line.
C. The graph of the equation is a circle with center
D. The radius of the circle is
D. The graph is nonexistent.

Ask by Sanders Ball. in the United States

Jan 21,2025

UpStudy AI · Accepted Answer

To determine whether the equation $$ x^{2} + y^{2} + 12x + 6y + 36 = 0 $$ represents a circle, we'll complete the square for both $$ x $$ and $$ y $$.

**Step 1: Rearrange the Equation**

$$
x^{2} + 12x + y^{2} + 6y = -36
$$

**Step 2: Complete the Square**

- **For $$ x $$:**
  
  $$
  x^{2} + 12x = (x + 6)^{2} - 36
  $$

- **For $$ y $$:**
  
  $$
  y^{2} + 6y = (y + 3)^{2} - 9
  $$

**Step 3: Substitute Back into the Equation**

$$
(x + 6)^{2} - 36 + (y + 3)^{2} - 9 = -36
$$

Combine the constants:

$$
(x + 6)^{2} + (y + 3)^{2} - 45 = -36
$$

$$
(x + 6)^{2} + (y + 3)^{2} = 9
$$

**Step 4: Interpret the Result**

The equation $$ (x + 6)^{2} + (y + 3)^{2} = 9 $$ is in the standard form of a circle:

$$
(x - h)^{2} + (y - k)^{2} = r^{2}
$$

where $$(h, k)$$ is the center and $$ r $$ is the radius.

- **Center $$(h, k)$$:** $$-6, -3$$
- **Radius $$ r $$:** $$\sqrt{9} = 3$$

**Conclusion:**

- **Option C:** The graph of the equation is a circle with center $$(-6, -3)$$.
- **Option D:** The radius of the circle is $$3$$.

**Final Answer:**

**C.** The graph of the equation is a circle with center $$(-6,\ -3)$$.

**D.** The radius of the circle is $$3$$.
The graph of the equation $$ x^{2} + y^{2} + 12x + 6y + 36 = 0 $$ is a circle with center at $$(-6, -3)$$ and a radius of $$3$$.

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