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

The correct option is C

$1$

Find the radius of the circle based on given information

Given, the plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$

Center of sphere is $\left(\frac{1}{2},0,\frac{-1}{2}\right)$

So, radius of sphere is $\sqrt{{\left(\frac{1}{2}\right)}^2+0^2+{\left(\frac{-1}{2}\right)}^2+2}=\sqrt{\frac{5}{2}}$

Now the perpendicular distance between centre of sphere of the given plane is given by,

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

Therefore, ${(\mathrm{Radius}\mathrm{of}\mathrm{circle})}^2={(\mathrm{Radius}\mathrm{of}\mathrm{sphere})}^2-{(\mathrm{Perpendicular}\mathrm{distance})}^2=\frac{5}{2}-\frac{9}{6}$

$Radiusofcircle=\sqrt{\frac{5}{2}-\frac{9}{6}}$

$$\begin{gathered} =\sqrt{\frac{30-18}{12}} \\ =\sqrt{\frac{12}{12}} \\ =1 \end{gathered}$$

Hence, option (C) is the correct answer.