Answer
$\text{Center: }
\left( -\dfrac{7}{2},-\dfrac{3}{2} \right)
\\\text{Radius: }
\dfrac{3\sqrt{6}}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the Center-Radius Form of the given equation, $
2x^2+2y^2+14x+6y+2=0
,$ complete the square for both $x$ and $y$ variables.
$\bf{\text{Solution Details:}}$
Grouping the $x$-variables and the $y$-variables and transposing the constant, the given equation is equivalent to
\begin{array}{l}\require{cancel}
(2x^2+14x)+(2y^2+6y)=-2
.\end{array}
Before completing the square, make the coefficient of $x^2$ and $y^2$ equal to $1.$ Hence, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( \dfrac{2}{2}x^2+\dfrac{14}{2}x \right)+\left( \dfrac{2}{2}y^2+\dfrac{6}{2}y \right)=-\dfrac{2}{2}
\\\\
\left( x^2+7x \right)+\left( y^2+3y \right)=-1
.\end{array}
Since the coefficient of $x$ is $
7
,$ add $\left(\dfrac{
7
}{2}\right)^2=
\dfrac{49}{4}
$ on both sides of the equation to complete the square for $x$. Hence, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( x^2+7x+\dfrac{49}{4} \right)+\left( y^2+3y \right)=-1+\dfrac{49}{4}
\\\\
\left( x+\dfrac{7}{2} \right)^2+\left( y^2+3y \right)=-\dfrac{4}{4}+\dfrac{49}{4}
\\\\
\left( x+\dfrac{7}{2} \right)^2+\left( y^2+3y \right)=\dfrac{45}{4}
.\end{array}
Since the coefficient of $y$ is $
3
,$ add $\left(\dfrac{
3
}{2}\right)^2=
\dfrac{9}{4}
$ on both sides of the equation to complete the square for $y$. Hence, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( x+\dfrac{7}{2} \right)^2+\left( y^2+3y+\dfrac{9}{4} \right)=\dfrac{45}{4}+\dfrac{9}{4}
\\\\
\left( x+\dfrac{7}{2} \right)^2+\left( y+\dfrac{3}{2} \right)^2=\dfrac{54}{4}
.\end{array}
Using the Center-Radius Form of the equation of circles which is given by $(x-h)^2+(y-k)^2=r^2,$ where the center is $(h,k)$ and the radius is $r,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( x-\left(-\dfrac{7}{2}\right) \right)^2+\left( y-\left(-\dfrac{3}{2} \right)\right)^2=\left(\sqrt{\dfrac{54}{4}}\right)^2
\\\\
\left( x-\left(-\dfrac{7}{2}\right) \right)^2+\left( y-\left(-\dfrac{3}{2} \right)\right)^2=\left(\sqrt{\dfrac{9}{4}\cdot6}\right)^2
\\\\
\left( x-\left(-\dfrac{7}{2}\right) \right)^2+\left( y-\left(-\dfrac{3}{2} \right)\right)^2=\left(\dfrac{3}{2}\sqrt{6}\right)^2
\\\\
\left( x-\left(-\dfrac{7}{2}\right) \right)^2+\left( y-\left(-\dfrac{3}{2} \right)\right)^2=\left(\dfrac{3\sqrt{6}}{2}\right)^2
.\end{array}
Hence the equation of the circle above has the following characteristics:
\begin{array}{l}\require{cancel}
\text{Center: }
\left( -\dfrac{7}{2},-\dfrac{3}{2} \right)
\\\text{Radius: }
\dfrac{3\sqrt{6}}{2}
.\end{array}
