## College Algebra (11th Edition)

$\text{Center: } \left( -\dfrac{7}{2},-\dfrac{3}{2} \right) \\\text{Radius: } \dfrac{3\sqrt{6}}{2}$
$\bf{\text{Solution Outline:}}$ 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}