question_answer

The values of\[\alpha \]for which the point\[(\alpha -1,\,\,\alpha +1)\]lies in the larger segment of the circle\[{{x}^{2}}+{{y}^{2}}-x-y-6=0\]made by the chord whose equation is\[x+y-2=0\]is

A) \[-1<\alpha <1\]                              

B) \[1<\alpha <\infty \]

C) \[-\infty <\alpha <-1\]  

D)        \[\alpha \le 0\]

Correct Answer: A

Solution :

The given circle                 \[S(x,\,\,y)\equiv {{x}^{2}}+{{y}^{2}}-x-y-6=0\]  ... (i) has centre at\[C\equiv \left( \frac{1}{2},\,\,\frac{1}{2} \right)\] According to the given conditions, the given point\[P(\alpha -1,\,\,\alpha +1)\]must lie inside the given circle. \[i.e.,\] \[S(\alpha -1,\,\,\alpha +1)<0\] \[\Rightarrow \]\[{{(\alpha -1)}^{2}}+{{(\alpha +1)}^{2}}-(\alpha -1)-(\alpha +1)-6<0\] \[\Rightarrow \]\[{{\alpha }^{2}}-\alpha -2<0\]\[i.e.,\]\[(\alpha -2)(\alpha +1)<0\] \[\Rightarrow \]               \[-1<\alpha <2\] [using sign - scheme from algebra]          ... (ii) and also\[P\]and\[C\]must lie on the same side of the line. \[L(x,\,y)\,=x+y-2=0\] \[i.e.,\]\[L\left( \frac{1}{2},\,\,\frac{1}{2} \right)\]and\[L(\alpha -1,\,\,\alpha +1)\]must have the same sign. Now, since\[L\left( \frac{1}{2},\,\,\frac{1}{2} \right)=\frac{1}{2}+\frac{1}{2}-2<0\] therefore, we have \[L(\alpha -1,\,\,\alpha +1)=(\alpha -1)+(\alpha +1)-2<0\] \[\Rightarrow \]               \[\alpha <1\]                                                      ... (iv) Ineqalities (ii) and (iv) together give the permissible values of\[\alpha \]as\[-1<\alpha <1\].