Question

The plane $2x+2y-z=k$ touches the sphere $x^2+y^2+z^2-4x+2y-6_Z+5=0$ and makes a positive intercept on the axis of $z$, then $k=$

• A. $-10$
• B. $-8$
• C. $18$
• D. $10$

Answer 1

The correct option is A $-10$

Centre and radius or the given sphere are $(2,-1,3)$ and $\sqrt{2^2+1^2+3^2-5}=3$
Since given plane touch the given sphere, so distance of plane from sphere's centre will be same as its radius.

$\Rightarrow \left|\frac{2\left(2\right)+2(-1)-\left(3\right)-k}{\sqrt{2^2+2^2+1^2}}\right|=3$

$\Rightarrow |k+1|=9\Rightarrow k=-10,8$
Since plane makes positive intercept on $z$-axis, hence, $k=-10$ is acceptable here.