Question

The Plane $x+y+z+k=0$ touches the sphere $x^2+y^2+z^2=k^2$ for $k\ne 0$

• A. All values of $k$
• B. Only one value of $k$
• C. Finite values of $k$
• D. No value of $k$

Answer 1

The correct option is D No value of $k$

Centre and radius of the given sphere are $(0,0,0)$ and $k$ respectively
Now the distance between centre of sphere to the given plane is given by,
$d=\left|\frac{k}{\sqrt{1^2+1^2+1^2}}\right|=\frac{k}{\sqrt{3}}\Rightarrow d<k$
Hence for any value of $k\ne 0$, given plane don't touch the sphere rather it cuts the sphere in to two parts.