A) 2
B) \[\sqrt{7}\]
C) 3
D) \[\sqrt{5}\]
Correct Answer: B
Solution :
Equation of sphere is, \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2y-4z=11\] Centre of sphere = (0, 1, 2) and radius of sphere = 4 Let centre of circle be\[(\alpha ,\beta ,\gamma )\] The d.r?s of line joining from centre of sphere to the centre of circle is \[(\alpha -0,\beta -1,\gamma -2)\]or \[(\alpha ,\beta -1,\gamma -2)\] But this line is normal at plane \[x+2y+2z=15\] \ \[\frac{\alpha }{1}=\frac{\beta -1}{2}=\frac{\gamma -2}{2}=k\] \[\alpha =k,\,\,\beta =2k+1,\gamma =2k+2\] \[\because \] Centre of circle lies on \[x+2y+2z=15\] \ \[k+2(2k+1)+2(2k+2)=15\]\[\Rightarrow k=1\] So, centre of circle = (1, 3, 4) Therefore, Radius of circle \[=\sqrt{{{(\text{Radius}\,\text{of}\,\text{sphere})}^{2}}-{{(\text{Length}\,\text{of}\,\text{joining}\,\text{line}\,\text{of}\,\text{centre})}^{2}}}\] \[=\sqrt{{{(4)}^{2}}-[{{(1-0)}^{2}}+{{(3-1)}^{2}}+{{(4-2)}^{2}}}]\]\[=\sqrt{16-9}=\sqrt{7}\].
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