Answer
$\text{Center: }
\left( -\dfrac{11}{2},\dfrac{5}{2} \right)
\\\text{Radius: }
\dfrac{\sqrt{146}}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}