Question

Find an equation of the sphere that passes through the origin and whose center is (-2, 2, 3). Be sure that your formula is monic. Equation: (x+2)^2+(y-2)^2+(z-3)^2 = 0

Answer 1

Answer:

$\bold{(x+2))^2+(y-2)^2+(z-2)^2=17}$

Step-by-step explanation:

Given the center of sphere is: (-2, 2, 3)

Passes through the origin i.e. (0, 0, 0)

To find:

The equation of the sphere ?

Solution:

First of all, let us have a look at the equation of a sphere:

$(x-a)^2+(y-b)^2+(z-c)^2=r^2$

Where ($x,y,z$) are the points on sphere.

$(a, b, c)$ is the center of the sphere and

$r$ is the radius of the sphere.

Radius of the sphere is nothing but the distance between any point on the sphere and the center.

We are given both the points, so we can use distance formula to find the radius of the given sphere:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Here,

$x_1 =0 \\y_1 =0 \\z_1 =0 \\x_2 =-2 \\y_2 =2 \\z_2 =3$

So, Radius is:

$r = \sqrt{(-2-0)^2+(2-0)^2+(3-0)^2}\\\Rightarrow r = \sqrt{4+4+9} = \sqrt{17}$

Therefore the equation of the sphere is:

$(x-(-2))^2+(y-2)^2+(z-2)^2=(\sqrt{17})^2\\\bold{(x+2))^2+(y-2)^2+(z-2)^2=17}$