Question

(X+2)^2 + (y-4)^2 = 0

Answer 1

(x+2)^2 + (y-4)^2 = 0
сумма двух неотрицательных чисел равна 0, каждое и них равно 0
(x+2)^2=0    x+2=0     x=-2
(y-4)^2=0     y-4=0      y=4

ответ (-2;4)

Answer 2

The equation you've provided is a conic section equation, specifically a circle equation. However, there seems to be a problem with the equation you've written because a circle equation should have positive constants on the right side, not equal to zero. The equation should typically be in the form:

(x - h)^2 + (y - k)^2 = r^2

where (h, k) represents the center of the circle, and r represents the radius of the circle. The equation you provided is not in this form, and the sum of squares on the left side should not be equal to zero for a valid circle equation.

If you have a specific question or if there was a typo in your equation, please provide more information, and I'd be happy to help you further.