Question

Find the sphere's center and radius. $$ 2 x^{2}+2 y^{2}+2 z^{2}-4 x-12 y-8 z+3=0 $$

Short answer

The center of the sphere is at the point (1, 3, 2) and the radius is \( \sqrt{\frac{1}{2}} \).

Step-by-step solution

Write the Sphere's Equation in Standard Form

Divide all terms in the given equation by 2 to simplify: \( x^{2} + y^{2} + z^{2} - 2x - 6y - 4z + \frac{3}{2} = 0 \). Now rearrange it to make squares: \( (x-1)^2 + (y-3)^2 + (z-2)^2 -\frac{1}{2} = 0 \). Adding \(\frac{1}{2}\) to both sides yields: \( (x-1)^2 + (y-3)^2 + (z-2)^2 = \frac{1}{2} \).

Identify the Center and Radius

Now the equation is in the form \( (x-a)^2 + (y-b)^2 + (z-c)^2 = r^2 \), where (a,b,c) is the center of the sphere, and r is the radius. Comparing this with the equation just derived, it’s clear that the center of the sphere is (1,3,2) and the radius of the sphere square is \(\frac{1}{2}\). Therefore, the radius of the sphere is \( \sqrt{\frac{1}{2}} \).