## Calculus (3rd Edition)

Write ${x^2} + 3{y^2} - 6x + 12y + 23 = 0$ ${x^2} - 6x + 3{y^2} + 12y + 23 = 0$ ${\left( {x - 3} \right)^2} - 9 + 3{\left( {y + 2} \right)^2} - 12 + 23 = 0$ ${\left( {x - 3} \right)^2} + 3{\left( {y + 2} \right)^2} + 2 = 0$ ${\left( {x - 3} \right)^2} + 3{\left( {y + 2} \right)^2} = - 2$ The left-hand side is nonnegative because it is the sum of two squares. But the right-hand side is negative, so there is no real solutions. Hence, it has no points.