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 $-1<\alpha <1$

The given circle $S(x,y)\equiv x^2+y^2-x-y-6=0$ has centre at $C\equiv (\frac{1}{2},\frac{1}{2})$
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]
and also $P$ and $C$ must lie on the same side of the line.
$L(x,y)=x+y-2=0$
i.e., $L(\frac{1}{2},\frac{1}{2})$ and $L(\alpha -1,\alpha +1)$ must have same sign.
Now, since $L(\frac{1}{2},\frac{1}{2})=\frac{1}{2}+\frac{1}{2}-2<0$
Therefore, we have
$L(\alpha -1,\alpha +1)=(\alpha -1)+(\alpha +1)-2<0$
$\Rightarrow \alpha <1$
Inequalities (ii) and (iv) together give the permissible values of $\alpha$ as $-1<\alpha <1$.