Question

The equation of the sphere concentric with the sphere $2x^2+2y^2+2z^2-6x+2y-4z=1$ and double its radius is:

• A. $x^2+y^2+z^2-x+y-z=1$
• B. $x^2+y^2+z^2-6x+2y-4z=1$
• C. $2x^2+2y^2+2z^2-6x+2y-4z-15=0$
• D. $x^2+y^2+z^2-3x+y-2z=1$
• E. $2x^2+2y^2+2z^2-6x+2y-4z-25=0$

Answer 1

Answer: (E)

Equation of sphere is $2x^2+2y^2+2z^2-6x-2y-4z-1=0$ Radius of sphere is
$\sqrt{\frac{9}{4}+\frac{1}{4}+\frac{4}{4}+\frac{1}{2}}=2$
Equation of family of concentric sphere is $x^2+y^2+z^2-3x+y-2z+\lambda =0$ ...(i)
$\therefore$ According to question, $\sqrt{\frac{9}{4}+\frac{1}{4}+1-\lambda }=4$
$\Rightarrow$ $\frac{14}{4}-\lambda =16$
$\Rightarrow$ $\lambda =\frac{14}{4}-16=-\frac{25}{2}$
$\therefore$ From Eq. (i)
$x^2+y^2+z^2-3x+y-2z-\frac{25}{2}=0$
$\Rightarrow$ $2x^2+2y^2+2z^2-6x+2y-4z-25=0$