the minimum distance between the circle x^2+y^2=9 and curve 2x^2+10y^2+6xy=1 is
A)2ROOT2 B)2 C)3-ROOT2 D)ROOT2

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$$\begin{gathered} \mathrm{These} \mathrm{type} \mathrm{of} \mathrm{questions} \mathrm{are} \mathrm{not} \mathrm{covered} \mathrm{in} \mathrm{IIT} \mathrm{JEE} \mathrm{syllabus}. \\ {\mathrm{x}}^2+{\mathrm{y}}^2=9..\left(\mathrm{i}\right) \\ \mathrm{Centre} \mathrm{of} \mathrm{cirlce} \mathrm{is} \left(0,0\right) \\ 2{\mathrm{x}}^2+10{\mathrm{y}}^2+6\mathrm{xy}=1...\left(\mathrm{ii}\right) \\ \mathrm{Partial} \mathrm{differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ 4\mathrm{x}+6\mathrm{y}=0 \\ \mathrm{Partial} \mathrm{differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{y} \\ 20\mathrm{y}+6\mathrm{x}=0 \\ \mathrm{Solution} \mathrm{of} \mathrm{above} \mathrm{gives} \left(\mathrm{x},\mathrm{y}\right)=\left(0,0\right) \\ \mathrm{Hence} \mathrm{centre} \mathrm{of} \left(\mathrm{ii}\right) \mathrm{is} \mathrm{also} \left(0,0\right) \\ \mathrm{Also} \mathrm{shortest} \mathrm{distance} \mathrm{is} \mathrm{always} \mathrm{found} \mathrm{along} \mathrm{common} \mathrm{normal}, \mathrm{and} \\ \mathrm{normal} \mathrm{of} \mathrm{circle} \mathrm{passes} \mathrm{through} \mathrm{centre} \left(0,0\right), \mathrm{hence} \mathrm{we} \mathrm{need} \mathrm{shortest} \\ \mathrm{distance} \mathrm{of} \mathrm{origin} \mathrm{from} \mathrm{any} \mathrm{point} \left(\mathrm{x},\mathrm{y}\right) \mathrm{on} \left(\mathrm{ii}\right). \\ \mathrm{Distance} \mathrm{of} \left(0,0\right) \mathrm{from} \left(\mathrm{x},\mathrm{y}\right) \mathrm{on} \left(\mathrm{ii}\right) \mathrm{is} \\ \mathrm{d}=\sqrt{{\left(\mathrm{x}-0\right)}^2+{\left(\mathrm{y}-0\right)}^2}=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2} \\ \Rightarrow {\mathrm{d}}^2={\mathrm{x}}^2+{\mathrm{y}}^2 \\ \mathrm{Now} \mathrm{you} \mathrm{you} \mathrm{to} \mathrm{minimize} {\mathrm{x}}^2+{\mathrm{y}}^2 \mathrm{subjected} \mathrm{to} \mathrm{condition} \\ 2{\mathrm{x}}^2+10{\mathrm{y}}^2+6\mathrm{xy}=1 \mathrm{which} \mathrm{will} \mathrm{give} \mathrm{d}=1 \\ \mathrm{Hence} \\ \mathrm{Shortest} \mathrm{distance}=\mathrm{Radius}-\mathrm{d}=3-1=2 \left[\mathrm{Answer}\right] \end{gathered}$$

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