A) 2
B) 1
C) 3
D) 0
Correct Answer: D
Solution :
the given circles are \[{{\left( x-y \right)}^{2}}+{{y}^{2}}=4\]and \[{{\left( x-1 \right)}^{2}}+{{y}^{2}}=16\] |
The points \[\left( a+1,\sqrt{3}a \right)\]lie on the line \[x=a+1,y=\sqrt{3}a\] |
\[i.e\,y=\sqrt{3}\left( x-y \right)\left[ e\operatorname{liminating}\,a \right]\] |
Whose slope =\[\sqrt{3,}\]hence makes angle \[60{}^\circ \]with the + ve direction of the x-axis. |
Hence, we have \[A\equiv (1+2\cos 60{}^\circ ,2\sin 60{}^\circ )\equiv (2,\sqrt{3})\] and \[B\equiv (1+4\cos 60{}^\circ ,4\sin 60{}^\circ )\equiv (3,2\sqrt{3})\] |
Hence, there is no point on the line segment \[AB\]whose abscissa is an integer since abscissa of \[A\]is 2 and that of \[B\]is 3. |
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