Question

1. in the xy - plane, the graph of $3x^{2}+18x + 3y^{2}+12y = 9$ is a circle. what is the centerpoint of the circle? (a) $(3,2)$ (b) $(6,4)$ (c) $(-6,-4)$ (d) $(-3,-2)$

Explanation:

Step1: Divide the equation by 3

$$\begin{align*} \frac{3x^{2}+18x + 3y^{2}+12y}{3}&=\frac{9}{3}\\ x^{2}+6x+y^{2}+4y& = 3 \end{align*}$$

Step2: Complete the square for \(x\) - terms

For \(x^{2}+6x\), using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\) and \(2ab = 6x\Rightarrow b = 3\). Then \(x^{2}+6x=(x + 3)^{2}-9\)

Step3: Complete the square for \(y\) - terms

For \(y^{2}+4y\), using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\) and \(2ab=4y\Rightarrow b = 2\). Then \(y^{2}+4y=(y + 2)^{2}-4\)

Step4: Substitute back into the equation

$$\begin{align*} (x + 3)^{2}-9+(y + 2)^{2}-4&=3\\ (x + 3)^{2}+(y + 2)^{2}&=3 + 9+4\\ (x + 3)^{2}+(y + 2)^{2}&=16 \end{align*}$$

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center of the circle.

Answer:

D. \((-3,-2)\)