Question

The locus of the equation $2(x^2+y^2+z^2)+2x-6y+4z+56=0$ is

• A. a sphere
• B. a degenerate sphere
• C. an empty set
• D. None of these

Answer 1

The correct option is C an empty set

$x^2+y^2+z^2+x-3y+2z+28=0$
${(x+\frac{1}{2})}^2+{(y-\frac{3}{2})}^2+{(z+1)}^2-\frac{1}{4}-\frac{9}{4}+28=0$
${(x+\frac{1}{2})}^2+{(y-\frac{3}{2})}^2+{(z+1)}^2-\frac{10}{4}+28=0$
${(x+\frac{1}{2})}^2+{(y-\frac{3}{2})}^2+{(z+1)}^2=\frac{-51}{2}$
Hence ${(x+\frac{1}{2})}^2+{(y-\frac{3}{2})}^2+{(z+1)}^2<0$
This is not possible for real values of $x$.
Hence the answer is the empty set.