Question

(x-3)^2+(y+2)^2+z^2=2​

Answer 1

Ответ:

1. x^2+y^2+z^2=49

Радиус =√49=7 Центр (0,0,0)

2. (x-3)^2+(y+2)^2+z^2=2

Радиус =√2 Центр (3,-2,0)

Пошаговое объяснение:

вроде так

Answer 2

The equation you've provided represents a three-dimensional geometric object, specifically a sphere. It is in the standard form for the equation of a sphere with its center at the point (3, -2, 0) and a radius of √2.

The general equation of a sphere in three-dimensional space is:

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

Where:

• (h, k, l) are the coordinates of the sphere's center.
• r is the radius of the sphere.

In your equation:

• (x - 3)^2 represents the (x - h)^2 term, where h = 3.
• (y + 2)^2 represents the (y - k)^2 term, where k = -2.
• z^2 represents the (z - l)^2 term, where l = 0.
• The right-hand side, 2, represents r^2, so r = √2.

So, the equation (x-3)^2 + (y+2)^2 + z^2 = 2 represents a sphere with its center at (3, -2, 0) and a radius of √2.