Question

the equation of a circle is given as $2x^2 + 2y^2 + 4x + 12y + 6 = 0$. what are the center, $c$, and the radius, $r$, of the circle?

$\circ$ $c(1, 3)$; $r = 7$
$\circ$ $c(-1, -3)$; $r = \sqrt{7}$
$\circ$ $c(-1, -3)$; $r = 7$
$\circ$ $c(1, 3)$; $r = \sqrt{7}$

Explanation:

Step1: Divide by 2 to simplify

Divide the entire equation \(2x^{2}+2y^{2}+4x + 12y+6 = 0\) by 2:
\(x^{2}+y^{2}+2x + 6y + 3=0\)

Step2: Group x and y terms

Group the \(x\)-terms and \(y\)-terms:
\((x^{2}+2x)+(y^{2}+6y)= - 3\)

Step3: Complete the square for x

For \(x^{2}+2x\), add \((\frac{2}{2})^{2}=1\) to both sides:
\((x^{2}+2x + 1)+(y^{2}+6y)=-3 + 1\)

Step4: Complete the square for y

For \(y^{2}+6y\), add \((\frac{6}{2})^{2}=9\) to both sides:
\((x^{2}+2x + 1)+(y^{2}+6y + 9)=-3 + 1+9\)

Step5: Write in standard form

Simplify to the standard circle equation \((x - h)^{2}+(y - k)^{2}=r^{2}\):
\((x + 1)^{2}+(y + 3)^{2}=7\)
Here, center \(C(h,k)=(-1,-3)\) and radius \(r=\sqrt{7}\) (since \(r^{2}=7\)).

Answer:

B. \(C(-1, -3)\); \(r = \sqrt{7}\) (assuming options are labeled as A, B, C, D with this option as B)