Solution 4.1:6c
From Förberedande kurs i matematik 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}