A) 0
B) 1
C) 2
D) None of these
Correct Answer: B
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 required conditions the given point \[P(\alpha -1,\alpha +1)\]Must lie inside the given circle | ||
i.e.\[S(\alpha -1\alpha +1)<0\operatorname{i}.e\] \[\]\[{{(\alpha -1)}^{2}}+{{(\alpha +1)}^{2}}-(\alpha -1)-(\alpha +1)-6<0\]\[\operatorname{i}.e.{{\alpha }^{2}}-\alpha -2<0\] | ||
\[i.e.(\alpha -2)-(\alpha +1)<0\] | ||
\[i.e.-1<\alpha <2\] | ? (ii) | |
And also \[P\]and \[C\]must lie on the same side of the line (see fig.) | ||
\[L(x,y)\equiv x+y-2=0\] | ? (iii) | |
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 \[L(\alpha -1,\alpha +1)\]\[=(\alpha -1)+(\alpha +1)-2<0\] | ||
i.e. \[\alpha <1\] | ?.(iv) | |
Inequalities (ii) and (iv) together give the permissible values of \[\alpha \]as \[\,-1<\alpha <1\] | ||
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