Question

The plane $2x-2y+z+12=0$ touches the sphere $x^2+y^2+z^2-2x-4y+2z-3=0$ at the point

• A. $(1,-4,-2)$
• B. $(-1,4,-2)$
  
• C. $(-1,-4.2)$
• D. $(1,4,-2)$

Answer 1

Answer: (B)

$(-1,4,-2)$

Equation of line passing through center of sphere $(1,2,-1)$ and perpendicular to the plane of sphere having direction ratios $(2,-2,1)$ is

$s:x^2+y^2+z^2-2x-4y+2z-3=0$

And plane $\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z+1}{1}=\lambda$

Any point on the line is $P(2\lambda +1,-2\lambda +2,\lambda -1)$

This point will lie in the plane, if it satisfies the given equation of the plane such that

$2(2\lambda +1)-2(-2\lambda +2)+(\lambda +1)+12=0\Rightarrow 9\lambda +9=0$

$\Rightarrow \lambda =-1$