Question

The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius

• A. $3$
• B. $1$
• C. $2$
• D. $\sqrt{2}$

Answer 1

Answer: (B)

$1$

Centre and radius of the given sphere are $(\frac{1}{2},0,\frac{-1}{2})$ and $\sqrt{{\left(\frac{1}{2}\right)}^2+0^2+{\left(\frac{-1}{2}\right)}^2+2}=\sqrt{\frac{5}{2}}$ respectively.
Now the perpendicular distance between centre of sphere to the given plane is given by,

$d=\left|\frac{1/2+2\left(0\right)+1/2-4}{\sqrt{1^2+2^2+1^2}}\right|=\frac{3}{\sqrt{6}}$

Therefore, radius of the circle in which the given sphere and plane cut is
$=\sqrt{\frac{5}{2}-\frac{9}{6}}=1$