• # question_answer The radius of the circle in which the sphere${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-x+z-2=0$cuts by the plane $x+2y-z=4,$is A)  3                B)  1 C)  2                D)  $\sqrt{2}$

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