The standard form of a circle is written as:

$(x-h)^2+(y-k)^2=r^2$

Write the following equation of a circle in standard form:

$3x^2-6x+3y^2-4y+2=0$

If the x² and y² coefficients are the same, the equation represents a circle. If the x² and y² coefficients are different, we may have an ellipse.

Divide both sides of the equation by 3 to reduce the x² and y² coefficients to 1.

$\frac{3x^2}{3}-\frac{6x}{3}+\frac{3y^2}{3}-\frac{4y}{3}+\frac{2}{3}=\frac{0}{3}$

$x^2-2x+y^2-\frac{4}{3}y+\frac{2}{3}=0$

Move $\frac{2}{3}$ to the right side of the equation.

$x^2-2x+y^2-\frac{4}{3}y=-\frac{2}{3}$

Complete the square for -2x.

$\left(\frac{-2}{2}\right)^2=\left(-1\right)^2=1$

$x^2-2x+\textbf{1}+y^2-\frac{4}{3}y=-\frac{2}{3}+\textbf{1}$

Complete the square for $-\frac{4}{3}y$.

$\left(\frac{-\frac{4}{3}}{2}\right)^2=\left(-\frac{4}{3}\left(\frac{1}{2}\right)\right)^2=\left(-\frac{4}{6}\right)^2=\left(-\frac{2}{3}\right)^2=\frac{4}{9}$

$x^2-2x+\textbf{1}+y^2-\frac{4}{3}y+\frac{\textbf{4}}{\textbf{9}}=-\frac{2}{3}+\textbf{1}+\frac{\textbf{4}}{\textbf{9}}$

$(x-1)(x-1)+(y-\frac{2}{3})(y-\frac{2}{3})=-\frac{6}{9}+\frac{9}{9}+\frac{4}{9}$

The equation for the circle in standard form is:

$(x-1)^2+(y-\frac{2}{3})^2=\frac{7}{9}$

The center of the circle is:

$(h,k)=(1, \frac{2}{3})$

The radius of the circle is:

$r=\sqrt{\frac{7}{9}}=\sqrt{\frac{1}{9}(7)}=\frac{1}{3}\sqrt{7}$