Question

a circle has the equation x² + y² + 3x + 6y + 9 = 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.

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for x and y terms.
For the x - terms:
\[x^{2}+3x=(x + \frac{3}{2})^{2}-\frac{9}{4}\]
For the y - terms:
\[y^{2}+6y=(y + 3)^{2}-9\]
The original equation \(x^{2}+y^{2}+3x + 6y+9 = 0\) becomes \((x+\frac{3}{2})^{2}-\frac{9}{4}+(y + 3)^{2}-9+9=0\), which simplifies to \((x+\frac{3}{2})^{2}+(y + 3)^{2}=\frac{9}{4}\).

Step2: Identify the center and radius

The standard form of a circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Comparing \((x+\frac{3}{2})^{2}+(y + 3)^{2}=\frac{9}{4}\) with the standard form, we have \(h=-\frac{3}{2}\), \(k=-3\) and \(r=\frac{3}{2}\).

Step3: Find the x - intercepts

Set \(y = 0\) in the original equation \(x^{2}+y^{2}+3x + 6y+9 = 0\), we get \(x^{2}+3x+9=0\).
The discriminant \(\Delta=b^{2}-4ac\) for the quadratic equation \(ax^{2}+bx + c=0\) (here \(a = 1\), \(b = 3\), \(c = 9\)) is \(\Delta=3^{2}-4\times1\times9=9 - 36=-27<0\), so there are no real x - intercepts.

Step4: Find the y - intercepts

Set \(x = 0\) in the original equation \(x^{2}+y^{2}+3x + 6y+9 = 0\), we get \(y^{2}+6y+9=0\).
Factor the quadratic equation: \((y + 3)^{2}=0\), so \(y=-3\).

Answer:

(a) Center \((h,k)=(-\frac{3}{2},-3)\), radius \(r=\frac{3}{2}\)
(b) To graph the circle, plot the center \((-\frac{3}{2},-3)\) and then draw a circle with radius \(\frac{3}{2}\) around it.
(c) The y - intercept is \(y=-3\), and there are no x - intercepts.