question_answer

The number of such points \[(a+1,\sqrt{3}a),\] where a is any integer, lying inside the region bounded by the circles\[{{x}^{2}}+{{y}^{2}}-2x-3=0\] and \[{{x}^{2}}+{{y}^{2}}-2x-15=0,\]

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.