Question

find the centre and radius of the sphere X square + Y square + Z square is equal to 1 , X + Y + Z is equal to zero

Answer 1

$x {}^{2} + y {}^{2} + z {}^{2} = 1$
$x + y + z = 0$

Answer 2

Answer:

Concept:

The collection of all points in three-dimensional Euclidean space R3 that are separated by r (also known as the "radius") from a certain point is known as a sphere (the "center"). The diameter is equal to twice the radius, and antipodes are pairs of locations on the sphere that are on opposing sides of a diameter. The typical sphere is a two-dimensional surface since the word "sphere" solely refers to the surface. Therefore, it is discouraged to use the term "sphere" in a colloquial sense to refer to a sphere's inside; instead, a ball is a more appropriate phrase.

Explanation:

Given:

$x^{2} +y^{2} +z^{2} - x -y -z$ = 1

$x + y +z$ = 0

Solution:

The center of sphere is

$x^{2} +y^{2} +z^{2} - x -y -z$ = 1

$x^{2} +y^{2} +z^{2} + 2ux + 2vy + 2wz + d = 0$

• $2ux = -x$

          $u = \frac{-1}{2}$

• $2vy = -y$

          $v = \frac{-1}{2}$

• $2wz = -z$

          $w = \frac{-1}{2}$

Centre = $( \frac{-1}{2} , \frac{-1}{2} , \frac{-1}{2} )$

The Radius of the sphere

Radius = $\sqrt{u^{2}+v^{2} +w^{2} -d }$

            = $\sqrt{(\frac{-1}{2})^{2} +(\frac{-1}{2})^{2} +(\frac{-1}{2})^{2} - 1 }$

            = $\sqrt{(\frac{1}{4}) + (\frac{1}{4})+ (\frac{1}{4}) - 1}$

            = $\sqrt{(\frac{3}{4}) - 1}$

            = $\sqrt{(\frac{2}{4})}$

            = $\sqrt{(\frac{1}{2})}$

#SPJ3