Aus Online Mathematik Brückenkurs 1

What we need to do is to rewrite the equation in the standard form

\displaystyle \left( x-a \right)^{2}+\left( y-b \right)^{2}=r^{2}

because then we can read off the circle's centre \displaystyle \left( a \right.,\left. b \right) and radius, \displaystyle r.

In our case, we need only take out the factor \displaystyle ~\text{3} from the brackets on the left-hand side

\displaystyle \begin{align} & \left( 3x-1 \right)^{2}+\left( 3y+7 \right)^{2}=3^{2}\left( x-\frac{1}{3} \right)^{2}+3^{2}\left( y+\frac{7}{3} \right)^{2} \\ & =9\left( x-\frac{1}{3} \right)^{2}+9\left( y+\frac{7}{3} \right)^{2} \\ \end{align}

and then divide both sides by \displaystyle \text{9} , so as to get the equation in the desired form:

\displaystyle \left( x-\frac{1}{3} \right)^{2}+\left( y+\frac{7}{3} \right)^{2}=\frac{10}{9}

Because the right-hand side can be written as \displaystyle \left( \sqrt{\frac{10}{9}} \right)^{2} and the term \displaystyle \left( y+\frac{7}{3} \right)^{2} as \displaystyle

\displaystyle \left( y-\left( -\frac{7}{3} \right) \right)^{2}, the equation describes a circle with its centre at \displaystyle \left( \frac{1}{3} \right.,\left. -\frac{7}{3} \right) and radius \displaystyle \sqrt{\frac{10}{9}}=\frac{\sqrt{10}}{3}