Answer:
[tex]\large \boxed{\sf \ \ 3 \ \ }[/tex]
Step-by-step explanation:
Hello, please consider the following.
[tex]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[/tex]
[tex]\text{*** simplify ***}\\\\\large \boxed{\s \ \ (x+2)^2+(y-\dfrac{3}{2})^2=\dfrac{9}{4} \ \ }[/tex]
So, the radius of the circle is...
[tex]\sqrt{\dfrac{9}{4}}=\dfrac{3}{2}[/tex]
..and the diameter is two times the radius, so this is is
[tex]\large \boxed{\sf \ \ 3 \ \ }[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
