## College Algebra (11th Edition)

$\text{Center: } \left( -\dfrac{11}{2},\dfrac{5}{2} \right) \\\text{Radius: } \dfrac{\sqrt{146}}{2}$
$\bf{\text{Solution Outline:}}$ To find the Center-Radius Form of the given equation, $3x^2+3y^2+33x-15y=0 ,$ complete the square for both $x$ and $y$ variables. $\bf{\text{Solution Details:}}$ Grouping the $x$-variables and the $y$-variables, the given equation is equivalent to \begin{array}{l}\require{cancel} (3x^2+33x)+(3y^2-15y)=0 .\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{3}{3}x^2+\dfrac{33}{3}x \right)+\left( \dfrac{3}{3}y^2-\dfrac{15}{3}y \right)=\dfrac{0}{3} \\\\ \left( x^2+11x \right)+\left( y^2-5y \right)=0 .\end{array} Since the coefficient of $x$ is $11 ,$ add $\left(\dfrac{ 11 }{2}\right)^2= \dfrac{121}{4}$ 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+11x+\dfrac{121}{4} \right)+\left( y^2-5y \right)=0+\dfrac{121}{4} \\\\ \left( x+\dfrac{11}{2} \right)^2+\left( y^2-5y \right)=\dfrac{121}{4} .\end{array} Since the coefficient of $y$ is $-5 ,$ add $\left(\dfrac{ -5 }{2}\right)^2= \dfrac{25}{4}$ 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{11}{2} \right)^2+\left( y^2-5y+\dfrac{25}{4} \right)=\dfrac{121}{4}+\dfrac{25}{4} \\\\ \left( x+\dfrac{11}{2} \right)^2+\left( y-\dfrac{5}{2} \right)^2=\dfrac{146}{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{11}{2}\right) \right)^2+\left( y-\left(\dfrac{5}{2}\right) \right)^2=\sqrt{\dfrac{146}{4}} \\\\ \left( x-\left(-\dfrac{11}{2}\right) \right)^2+\left( y-\left(\dfrac{5}{2}\right) \right)^2=\sqrt{\dfrac{1}{4}\cdot146} \\\\ \left( x-\left(-\dfrac{11}{2}\right) \right)^2+\left( y-\left(\dfrac{5}{2}\right) \right)^2=\dfrac{1}{2}\sqrt{146} \\\\ \left( x-\left(-\dfrac{11}{2}\right) \right)^2+\left( y-\left(\dfrac{5}{2}\right) \right)^2=\dfrac{\sqrt{146}}{2} .\end{array} Hence the equation of the circle above has the following characteristics: \begin{array}{l}\require{cancel} \text{Center: } \left( -\dfrac{11}{2},\dfrac{5}{2} \right) \\\text{Radius: } \dfrac{\sqrt{146}}{2} .\end{array}