question_answer

The number of points with integral coordinates that lie in the interior of the region common to the circle \[{{x}^{2}}+{{y}^{2}}=16\] and the parabola \[{{y}^{2}}=4x\], is

A) \[8\]                                     

B) \[10\]

C) \[16\]                                   

D)  None of these

Correct Answer: A

Solution :

Let \[(\alpha ,\,\,\beta )\] be the point with integral coordinates and lying in the interior of the region common to the circle \[{{x}^{2}}+{{y}^{2}}=16\] and the parabola\[{{y}^{2}}=4x\]. Then,                    \[{{\alpha }^{2}}+{{\beta }^{2}}-16<0\] and        \[{{\beta }^{2}}-4\alpha <0\] It is clear from the figure that                 \[0<\alpha <4\] \[\Rightarrow \]               \[\alpha =1,\,\,2,\,\,3\]                 \[[\because \,\,\alpha \in Z]\] When    \[\alpha =1\]                 \[{{\beta }^{2}}<4\alpha \] \[\Rightarrow \]               \[{{\beta }^{2}}<4\] \[\Rightarrow \]               \[\beta =0,\,\,1\] So, the points are \[(1,\,\,0)\] and\[(1,\,\,1)\]. When    \[\alpha =2\]                 \[{{\beta }^{2}}<4\alpha \] \[\Rightarrow \]               \[{{\beta }^{2}}<8\] \[\Rightarrow \]               \[\beta =0,\,\,1,\,\,2\] So, the points are \[(2,\,\,0),\,\,(2,\,\,1)\] and\[(2,\,\,2)\]. When    \[\alpha =3\]                 \[{{\beta }^{2}}<4\alpha \] \[\Rightarrow \]               \[{{\beta }^{2}}<12\] \[\Rightarrow \]               \[\beta =0,\,\,1,\,\,2,\,\,3\] So, the points are\[(3,\,\,0),\,\,(3,\,\,1),\,\,(3,\,\,2)\]and\[(3,\,\,3)\]. Out of these points, \[(3,\,\,3)\] does not satisfy\[{{\alpha }^{2}}+{{\beta }^{2}}-16<0\]. Thus, the points lying in the region are \[(1,\,\,0),\,\,(1,\,\,1),\,\,(2,\,\,0),\,\,(2,\,\,1),\,\,(2,\,\,2),\,\,(3,\,\,0),\,\,(3,\,\,1)\]and (3,2).