Question

the equation of a circle is given in general form. write the equation of the circle in standard form. 3x² + 3y² - 6x + 12y + 3 = 0 the standard form of the equation is . (simplify your answer.)

Explanation:

Step1: Divide by coefficient of squared terms

Divide the entire equation $3x^{2}+3y^{2}-6x + 12y+3 = 0$ by 3. We get $x^{2}+y^{2}-2x + 4y + 1=0$.

Step2: Rearrange terms

Group the $x$ - terms and $y$ - terms together: $(x^{2}-2x)+(y^{2}+4y)=-1$.

Step3: Complete the square for $x$ - terms

For the $x$ - terms $x^{2}-2x$, we add $(\frac{-2}{2})^{2}=1$ to both sides of the equation. For the $y$ - terms $y^{2}+4y$, we add $(\frac{4}{2})^{2}=4$ to both sides of the equation. So, $(x^{2}-2x + 1)+(y^{2}+4y + 4)=-1+1 + 4$.

Step4: Write in standard form

The standard - form of a circle is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center of the circle and $r$ is the radius. The equation $(x - 1)^{2}+(y + 2)^{2}=4$ is in standard form.

Answer:

$(x - 1)^{2}+(y + 2)^{2}=4$