Question

(x-3)²+(y-5)²=0(x²-4)²+(y+1)²=0

Answer 1

1) х=3; у=5
2) х=2 ; у=-1

Answer 2

Let's analyze the given equations:

1. (x - 3)² + (y - 5)² = 0
2. (x² - 4)² + (y + 1)² = 0

Both equations have the form of a circle equation, which is (x - h)² + (y - k)² = r², where (h, k) represents the center of the circle, and r represents the radius.

Now, for a circle equation, if the radius (r) is 0, then it represents a single point. Let's check if that's the case for these two equations:

1. (x - 3)² + (y - 5)² = 0: The equation represents a circle centered at (3, 5) with a radius of 0. This means that the only solution to this equation is the point (3, 5).

2. (x² - 4)² + (y + 1)² = 0: The equation represents a circle centered at (0, -1) with a radius of 0. This means that the only solution to this equation is the point (0, -1).

So, both equations only have a single solution, which is a point, and that point is (3, 5) for the first equation and (0, -1) for the second equation.