A circle has the equation x squared plus y squared plus 3 x plus 6 y plus 9 equals 0.
​(a) Find the center​ (h,k) and radius r of the circle.
​(b) Graph the circle.
​(c) Find the​ intercepts, if​ any, of the graph.

Let's tackle the given equation of the circle step by step.

### Given equation:
\[
x^2 + y^2 + 3x + 6y + 9 = 0
\]

We will need to convert this into the standard form of a circle equation, which is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \( (h, k) \) is the center of the circle, and \( r \) is the radius.

### (a) Find the center \((h,k)\) and radius \(r\).

#### Step 1: Complete the square

We will complete the square for both \(x\) and \(y\) terms.

1. **Group the \(x\) terms and the \(y\) terms:**
   \[
   (x^2 + 3x) + (y^2 + 6y) = -9
   \]

2. **Complete the square for \(x^2 + 3x\):**
   To complete the square, take half the coefficient of \(x\), square it, and add/subtract it:
   \[
   \left(\frac{3}{2}\right)^2 = \frac{9}{4}
   \]
   So we add \(\frac{9}{4}\) to both sides:
   \[
   (x^2 + 3x + \frac{9}{4}) = -9 + \frac{9}{4}
   \]

3. **Complete the square for \(y^2 + 6y\):**
   Similarly, take half the coefficient of \(y\), square it, and add/subtract it:
   \[
   \left(\frac{6}{2}\right)^2 = 9
   \]
   Add \(9\) to both sides:
   \[
   (y^2 + 6y + 9) = -9 + \frac{9}{4} + 9
   \]

#### Step 2: Simplify the equation
\[
(x + \frac{3}{2})^2 + (y + 3)^2 = \frac{9}{4}
\]

Now, this is in standard form:
\[
\left(x + \frac{3}{2}\right)^2 + (y + 3)^2 = \left(\frac{3}{2}\right)^2
\]

- The center \((h, k)\) is:
  \[
  h = -\frac{3}{2}, \, k = -3
  \]
- The radius \(r\) is:
  \[
  r = \frac{3}{2}
  \]

### (b) Graph the circle

To graph the circle:
- The center is at \( \left( -\frac{3}{2}, -3 \right) \).
- The radius is \( \frac{3}{2} \), so the circle has a radius of \( 1.5 \).

You can sketch this by plotting the center and drawing a circle with radius 1.5 around it.

### (c) Find the intercepts

#### x-intercepts:
Set \( y = 0 \) in the original equation:
\[
x^2 + 3x + 0^2 + 6(0) + 9 = 0
\]
\[
x^2 + 3x + 9 = 0
\]
The discriminant of this quadratic equation is:
\[
\Delta = b^2 - 4ac = 3^2 - 4(1)(9) = 9 - 36 = -27
\]
Since the discriminant is negative, there are no real solutions for \(x\), so there are no x-intercepts.

#### y-intercepts:
Set \( x = 0 \) in the original equation:
\[
0^2 + y^2 + 3(0) + 6y + 9 = 0
\]
\[
y^2 + 6y + 9 = 0
\]
\[
(y + 3)^2 = 0
\]
\[
y = -3
\]
So, the only y-intercept is \( y = -3 \).

### Summary:
- The center is \( \left( -\frac{3}{2}, -3 \right) \).
- The radius is \( \frac{3}{2} \).
- There are no x-intercepts.
- The y-intercept is \( (0, -3) \).

Would you like a more detailed graph, or do you have any further questions?

---

Here are five related questions to practice:
1. How would the equation change if the center was shifted to \( (2, -4) \)?
2. What happens to the radius if the constant term changes in the equation?
3. How do you find the area of the circle with radius \( \frac{3}{2} \)?
4. What are the general steps for completing the square in any quadratic equation?
5. Can a circle have more than two intercepts? Why or why not?

**Tip:** Always check the discriminant to determine whether a quadratic equation has real solutions when finding intercepts.

A circle has the equation x squared plus y squared plus 3 x plus 6 y plus 9 equals 0. (a) Find the center​ (h,k) and radius r of the circle. ​(b) Graph the circle. ​(c) Find the​ intercepts, if​ any, of the graph.

Learn how to find the center, radius, and intercepts of a circle from the equation x² + y² + 3x + 6y + 9 = 0. This problem involves completing the square and understanding the discriminant of quadratic equations, suitable for students in Grades 9-11.

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