Answer:
[tex]\alpha=3[/tex]
Step-by-step explanation:
Equation of a Circle
A circle of radius r and centered on the point (h,k) can be expressed by the equation
[tex](x-h)^2+(y-k)^2=r^2[/tex]
We are given the equation of a circle as
[tex]3x^2+3y^2-6\alpha x+12y-3\alpha=0[/tex]
Note we have corrected it by adding the square to the y. Simplify by 3
[tex]x^2+y^2-2\alpha x+4y-\alpha=0[/tex]
Complete squares and rearrange:
[tex]x^2-2\alpha x+y^2+4y=\alpha[/tex]
[tex]x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4[/tex]
[tex](x-\alpha)^2+(y+2)^2=r^2[/tex]
We can see that, if r=4, then
[tex]\alpha+\alpha^2+4=16[/tex]
Or, equivalently
[tex]\alpha^2+\alpha-12=0[/tex]
There are two solutions for [tex]\alpha[/tex]:
[tex]\alpha=-4,\ \alpha=3[/tex]
Keeping the positive solution, as required:
[tex]\boxed{\alpha=3}[/tex]
