Question

Show that the equation represent a sphere, and find its center and radius. 3x2+3y2+3z2=10+6y+12z

Answer 1

Answer:

Center of sphere = (0, 1, 2)

Radius of sphere = $\dfrac{5}{\sqrt3}\text{ units}$

Step-by-step explanation:

We are given the following in the question:

Equation of sphere:

$3x^2+3y^2+3z^2=10+6y+12z$

Formula:

The equation of sphere is of the form

$(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\\\text{where (a,b,c) is the centre of sphere and r is the radius of sphere.}$

Simplifying the given equation we get,

$3x^2+3y^2+3z^2=10+6y+12z\\\text{Dividing by 3}\\x^2 + y^2 + z^2 = \dfrac{10}{3} + 2y + 4z\\\\x^2 + y^2 -2y + z^2-4z = \dfrac{10}{3}\\\\\text{Adding 1 and 4 on both sides, we get,}\\\\x^2 + y^2 -2y +1 + z^2-4z + 4 = \dfrac{10}{3} + 1+ 4\\\\(x-0)^2 + (y-1)^2 + (z-2)^2 = \dfrac{25}{3}\\\\\text{Comparing with the equation of sphere}\\a = 0\\b = 1\\c = 2\\\\r^2 = \dfrac{25}{3}\\\\r = \sqrt{\dfrac{25}{3}}= \dfrac{5}{\sqrt3}$

Center of sphere = (0, 1, 2)

Radius of sphere = $\dfrac{5}{\sqrt3}\text{ units}$