Question

(х-2)² + (у+3)² + z²=25

Answer 1

(x-2)²+(y+3)²+z²=25
(x-2)²+(y+3)²+z²=5² - уравнение сферы с центром в точке А(2;-3;0) и радиусом R=5

Answer 2

The equation (х-2)² + (у+3)² + z²=25 represents a sphere centered at the point (2,-3,0) with a radius of 5.

To see why this is the case, let's first rewrite the equation in standard form:

(х-2)² + (у+3)² + z² = 5²

We can see that this equation is in the standard form of the equation of a sphere, which is:

(x - a)² + (y - b)² + (z - c)² = r²

where (a, b, c) is the center of the sphere and r is the radius.

So in this case, we can identify the center of the sphere as (2, -3, 0) and the radius as 5. This means that any point (x, y, z) that satisfies the equation is on the surface of the sphere, and any point that is not on the sphere does not satisfy the equation.

For example, the point (2, -3, 0) satisfies the equation because:

(2-2)² + (-3+3)² + 0² = 5²

0 + 0 + 0 = 25

So it is on the surface of the sphere. However, the point (2, -8, 0) does not satisfy the equation because:

(2-2)² + (-8+3)² + 0² = 5²

0 + 25 + 0 = 25

So it is not on the surface of the sphere.