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

Answer 1

Answer: (C)

Since the centre of sphere
$x^{2}+y^{2}+z^{2}-x+z-2=0$ is $\left(\frac{1}{2}, 0, -\frac{1}{2}\right)$ and radius of sphere
$=\sqrt{\frac{1}{4}+\frac{1}{4}+2}=\frac{\sqrt{10}}{2}$
Distance of plane from centre of sphere
$=\left|\frac{\frac{1}{2}+\frac{1}{2}-4}{\sqrt{1+4+1}}\right|=\frac{3}{\sqrt{6}}$
So radius of circle $=\sqrt{\frac{10}{4}-\frac{9}{6}}$
$=\sqrt{\frac{30-18}{12}}=\sqrt{\frac{12}{12}}=1$