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  • The number of integral values of    for which the point <img alt="(\alpha-1, \alpha+1)" src="h

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

Option: 1

1


Option: 2

-1


Option: 3

0


Option: 4

-2


Answers (1)

best_answer

\mathrm{S(x, y)=x^2+y^2-x-y-6=0}.....................(i)
has centre at \mathrm{C \equiv\left(\frac{1}{2}, \frac{1}{2}\right)}
According to the required conditions, the given point \mathrm{P(\alpha-1, \alpha+1)}  must lie inside the given circle.
i.e. \mathrm{S(\alpha-1, \alpha+1)<0}

\mathrm{ \Rightarrow(\alpha-1)^2+(\alpha+1)^2-(\alpha-1)-(\alpha+1)-6<0 \\ }

\mathrm{ \Rightarrow \alpha^2-\alpha-2<0 }
\mathrm{ \Rightarrow \quad(\alpha-2)(\alpha+1)<0 \\ }

\mathrm{ \Rightarrow \quad-1<\alpha<2 }...........................(ii)

Also, P and C must lie on the same side of the line

\mathrm{ L(x, y) \equiv x+y-2=0 }..................(iii)
i.e. \mathrm{ L(1 / 2,1 / 2)~ and ~L(\alpha-1, \alpha+1) }   must have the same sign.
Since \mathrm{ L\left(\frac{1}{2}, \frac{1}{2}\right)=\frac{1}{2}+\frac{1}{2}-2<0 }
\mathrm{ \therefore \quad L(\alpha-1, \alpha+1)=(\alpha-1)+(\alpha+1)-2<0}
i.e.,\mathrm{ \alpha<1}                         .............................(iv)
Inequalities (ii) and (iv) together give the permissible values of \mathrm{ \alpha}  as  \mathrm{ -1<\alpha<1}

Posted by

sudhir.kumar

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