A) \[A\subset B\]
B) \[B\subset A\]
C) \[A=B\]
D) None of these
Correct Answer: A
Solution :
We have, \[|x-3|<1\]and\[|y-3|<1\] \[\Rightarrow \] \[2<x<4\]and\[2<y<4\] Thus, \[A\] is the set of all points \[(x,\,\,y)\] lying inside the square formed by the lines\[x=2,\,\,x=4,\,\,y=2\] and\[y=4\]. Now, \[4{{x}^{2}}+9{{y}^{2}}-32x-54y+109\le 0\] \[\Rightarrow \] \[4({{x}^{2}}-8x)+9({{y}^{2}}-6y)+109\le 0\] \[\Rightarrow \] \[4{{(x-4)}^{2}}+9{{(y-3)}^{2}}\le 36\] \[\Rightarrow \] \[\frac{{{(x-4)}^{2}}}{{{3}^{2}}}+\frac{{{(y-3)}^{2}}}{{{2}^{2}}}\le 1\] Thus, \[B\] is the set of all points lying inside the ellipse having its centre at \[(4,\,\,3)\] and of lengths major and minor axes are 3 and 2 units. It can be easily seen by drawing the graphs of two regions that\[A\subset B\].
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