Question

(+25 points!) Completeing the square.... Please help meee:)

Answer 1

Answer:

$\large \boxed{\sf \ \ 3 \ \ }$

Step-by-step explanation:

Hello, please consider the following.

$3x^2+3y^2+12x-9y+12=0\\\\\text{*** divide by 3 both sides***}\\\\x^2+y^2+4x-3y+4=0\\\\\text{*** Let's notice that: ***}\\\\*** \ x^2+4x=(x+2)^2-2^2=(x+2)^2-4 \ \ and \\\\*** \ y^2-3y=(y-\dfrac{3}{2})^2-\dfrac{3^2}{2^2}=(y-\dfrac{3}{2})^2-\dfrac{9}{4}\\\\\text{ *** complete the two squares ***}\\\\x^2+y^2+4x-3y+4=(x+2)^2-4+(y-\dfrac{3}{2})^2-\dfrac{9}{4}+4=0$

$\text{*** simplify ***}\\\\\large \boxed{\s \ \ (x+2)^2+(y-\dfrac{3}{2})^2=\dfrac{9}{4} \ \ }$

So, the radius of the circle is...

   $\sqrt{\dfrac{9}{4}}=\dfrac{3}{2}$

..and the diameter is two times the radius, so this is is

$\large \boxed{\sf \ \ 3 \ \ }$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you