Question

The equation of sphere whose centre is (2,-3,-1) and radius 4 is
Options
O (x - 2)2 + y + 3)2 + (z + 1)2 = 16
O (x - 2)2 + (y - 3)2 + (z - 1)2 = 16
O (x - 2)2 + (y - 3)2 + (z + 1)= 16
O (x + 2)2 + (y + 3)2 + (z + 1)2 = 16​

Answer 1

1 - (x-2)² + (y+3)² + (z+1)² = 16.

Step-by-step explanation:

The general equation of the sphere is

$x2 + y2 + z2 = r2$

The general equation when centre is given is :

$(x−h) {}^{2} +(y−k) {}^{2} +(z−l) {}^{2} =r {}^{2}$

in the question given :

centre : (2,-3,-1)

1. h= 2
2. k= -3
3. l= -1
4. The given radius is (r) : 4units.

Therefore the equation will be :

(x-2)² + (y-(-3))² + (z-(-1))² = 4²

(x-2)² + (y+3)² + (z+1)² = 16.

hence , (x-2)² + (y+3)² + (z+1)² = 16.

(#SPJ3)

Answer 2

Answer:

The answer is option [A]

$(x-2)^{2} + (y+3)^{2} + (z+1)^{2} = 16$

step-by-step explanation:

The x, y, and z axes are the three axes that describe a sphere. A circle's circumference defines an area. The area's equation is $r^2$. The surface area of the outer surface of a sphere is equivalent to $4r^2$. There isn't any volume.

A sphere's basic formula is $(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2,$ where (a, b, c) denotes the sphere's center, r denotes its radius, and x, y, and z denote the points that are located on the sphere's surface.

The sphere's general equation is

$x^2+y^2+z^2 = r^2$

When a center is provided, the generic equation is:

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

$center : (2,-3,-1)$

$h= 2$

$k= -3$

$l= -1$

The given radius is (r) $: 4units$

Consequently, this equation will be:

$(x-2)^2 + (y-(-3))^2 + (z-(-1))^2 = 4^2$

$(x-2)^2 + (y+3)^2 + (z+1)^2 = 16$

Therefore, the answer is

$(x-2)^2 + (y+3)^2 + (z+1)^2 = 16$

#SPJ2