Question

The circle $x^2+y^2+z^2+2x-4y+6z-8=0$, $x+y+z=1$ is a greatest circle of the sphere

• A. $x^2+y^2+z^2+2x-4y+6z-8=0$
• B. $x^2+y^2+z^2+3x-3y+5z-9=0$
• C. $x^2+y^2+z^2-6y+4z-6=0$
• D. None of these

Answer 1

The correct option is D None of these

Required sphere is given by,
$x^2+y^2+z^2+2x-4y+6z-8+\lambda (x+y+z-1)=0$
$\Rightarrow x^2+y^2+z^2+(2+\lambda )x+(\lambda -4)y+(6+\lambda )z-(\lambda +8)=0$
Thus centre of the sphere is, $(1+\frac{\lambda }{2},\frac{\lambda }{2}-2,3+\frac{\lambda }{2})$
Since given circle is greatest circle of this sphere so the centre will lie in the plane $x+y+z=1$
$\Rightarrow 2+\frac{3\lambda }{2}=1\Rightarrow \lambda =-\frac{2}{3}$
Hence sphere is $3(x^2+y^2+z^2)+4x-14y+16z-22=0$