Question 98840
Well, your error is found in the step..."completing the square"
Let's go over that:
{{{3x^2-6x+3y^2+12y = 0}}} then you said ..."complete the square", but before you do that, you should factor out a 3 then divide through by 3 and you will get:
{{{3(x^2-2x+y^2+4) = 0}}} Divide both sides by 3.
{{{x^2-2x+y^2+4y = 0}}} Now you complete the square when the coefficients of the {{{x^2}}} and {{{y^2}}} terms are equal to 1.
{{{(x^2-2x+1)+(y^2+4y+4) = 1+4}}} Factor the trinomials.
{{{(x-1)^2+(y+2)^2 = 5}}} Now compare this with the standard form for a circle of radius r and center at (h, k).
{{{(x-h)^2+(y-k)^2 = r^2}}}and you can see that h = 1, k = -2, and {{{r^2 = 5}}}
The center (h, k) is at (1, -2) and the radius is {{{r = sqrt(5)}}}