Question

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. $\alpha \le 0$

Answer 1

The correct option is A $-1<\alpha <1$

Circle $S=x^2+y^2-x-y-6=0$
Its centre is at $C(\frac{1}{2},\frac{1}{2})$
$P(\alpha -1,\alpha +1)$ must lie inside the circle.
So, $(\alpha -1{)}^2+(\alpha +1{)}^2-(\alpha -1)-(\alpha +1)-6<0$
$\Rightarrow {\alpha }^2-\alpha -2<0\Rightarrow (\alpha -2)(\alpha +1)<0$
$\Rightarrow -1<\alpha <2$ ... (1)
Also, P and C must lie on the same side of the line
$L=x+y-2=0$
That is, $L(\frac{1}{2},\frac{1}{2})$ and $L(\alpha -1,\alpha +1)$ must have same sign.
$L(\frac{1}{2},\frac{1}{2})$
$=\frac{1}{2}+\frac{1}{2}-2=-1<0$
$L(\alpha -1,\alpha +1)=\alpha -1+\alpha +1-2<0$
$\Rightarrow \alpha <1$ ... (2)
From (1) and (2), $-1<\alpha <1$