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?

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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.