Question 48282
I am only guessing that this is equal to zero.
2x^2 + y^2 - 4x - 4y = 0
2x^2 - 4x + y^2 - 4y = 0
2(x^2 - 2x) + (y^2 - 4y) = 0
2(x - 1)^2 + (y - 2)^2 = 0 + 4 + 2 = 6
(x - 1)^2/3 + (y - 2)^2/6 = 1
{{{graph(600,600,-10,10,-10,10,sqrt(6 - 2(x - 1)^2)+2,-sqrt(6 - 2(x - 1)^2)+2)}}}
Intercept: P(0,4) .... P(0,0) .... P(2,0)