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

Answer: (A)

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 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] ... (ii)
and also $P$ and $C$ must lie on the same side of the line.

$L(x,y)=x+y-2=0$
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
$L(\alpha -1,\alpha +1)=(\alpha -1)+(\alpha +1)-2<0$
$\Rightarrow \alpha <1$ ... (iv)
Ineqalities (ii) and (iv) together give the permissible values of $\alpha$ as $-1<\alpha <1$.