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 $\sqrt{\frac{1}{4}+\frac{1}{4}+2}=\sqrt{\frac{5}{2}}$ respectively.
Now perpendicular distance of centre from plane $x+2y-z=4$ is,
$=\left|\frac{\frac{1}{2}+\frac{1}{2}-4}{\sqrt{6}}\right|=\sqrt{\frac{3}{2}}$
Hence, radius of the circle is $=\sqrt{\frac{5}{2}-\frac{3}{2}}=1$.