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

Option: 1

-1<\alpha<1


Option: 2

1<\alpha<\infty


Option: 3

-\infty<\alpha<-1


Option: 4

\alpha \leq 0


Answers (1)

The given circle \mathrm{S(x, y)=x^{2}+y^{2}-x-y-6=0 \quad \ldots(i)}

\mathrm{Centre \: \mathrm{C}=\left(\frac{1}{2}, \frac{1}{2}\right)}
The given point \mathrm{\mathrm{P}(\alpha-1, \alpha+1)} must lie inside the given circle
i.e. \mathrm{\mathrm{S}(\alpha-1, \alpha+1)<0 i.e., (\alpha-1)^{2}+(\alpha+1)^{2}-(\alpha-1)-(\alpha+1)-6<0}

\mathrm{i.e., \alpha^{2}-\alpha-2<0 i.e., (\alpha-2)(\alpha-1)<0}

\mathrm{i.e., -1<\alpha<2\: \quad \ldots{(ii)}}

\mathrm{\mathrm{P}} must lie same side as the centre lies of the line \mathrm{\mathrm{x}+\mathrm{y}-2=0}
\mathrm{i.e., \mathrm{L}\left(\frac{1}{2}, \frac{1}{2}\right)\: and \: \mathrm{L}(\alpha-1, \alpha+1)}

\mathrm{Now \quad \mathrm{L}\left(\frac{1}{2}, \frac{1}{2}\right)=\frac{1}{2}+\frac{1}{2}-2<0}
\mathrm{\therefore \quad \mathrm{L}(\alpha-1), \alpha+1)=(\alpha-1)+(\alpha+1)-2<0 i.e., \alpha<1 \quad \ldots (iii)}

In equalities (ii) and (iii) together give the permissible values of  \mathrm{\alpha\, as -1<\alpha<1}.
 

Posted by

Ramraj Saini

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