Mathematics - Circle Question with Solution | TestHub

$\mathrm{S}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}-\mathrm{y}-6=0$

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. $\mathrm{S}(\alpha -1,\alpha +1)<0$

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

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

$\Rightarrow -1<\alpha <2$

Also P and C must lie on the same side of the line (see figure)

i.e. $\mathrm{L}(1/2,1/2)$ and $\mathrm{L}(\alpha -1,\alpha +1)$ must have the same sign.

Since $\mathrm{L}\left(\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}+\frac{1}{2}-2<0$

$\mathrm{L}(\alpha -1,\alpha +1)=(\alpha -1)+(\alpha +1)-2<0$, i.e., $\alpha <1$.........(4)

Inequalities (2) and (4) together give the permissible values of $\alpha$ as $-1<\alpha <1$.