A) 3
B) 1
C) 2
D) \[\sqrt{2}\]
Correct Answer: B
Solution :
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\[=\sqrt{\frac{10}{4}-\frac{9}{6}}=\sqrt{\frac{30-18}{12}}\] \[=1\]
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