Question

If the point $(-5+\frac{\lambda }{\sqrt{2}},-3+\frac{\lambda }{\sqrt{2}})$ is an interior point of the larger segment of the circles $x^2+y^2=16$ cut off by the line $x+y=2$, then

• A. $\lambda ϵ(-\infty ,5\sqrt{2})$
• B. $\lambda ϵ(4\sqrt{2}-\sqrt{14},5\sqrt{2})$
• C. $\lambda ϵ(4\sqrt{2}-\sqrt{14},4\sqrt{2}+\sqrt{14})$
• D. None of these

Answer 1

The correct option is B $\lambda ϵ(4\sqrt{2}-\sqrt{14},5\sqrt{2})$

Hence, the larger segment of circle lies towards the side having origin with respect to the line

$x+y-2=0$

For origin $\to 0+0-2=-2<0$

For the given point:

$\Rightarrow -5+\frac{\lambda }{\sqrt{2}}-3+\frac{\lambda }{\sqrt{2}}-2<0$

$\Rightarrow -10+\lambda \sqrt{2}<0$

$\Rightarrow \lambda \sqrt{2}<10$

$\Rightarrow \lambda <\frac{10}{\sqrt{2}}$

$\Rightarrow \lambda <5\sqrt{2}\to \left(1\right)$

Also this point is the interior point of circle.

$\Rightarrow (-5+\frac{\lambda }{\sqrt{2}}{)}^2+(-3+\frac{\lambda }{\sqrt{2}}{)}^2-16<0$

$\Rightarrow 25+\frac{{\lambda }^2}{2}+9+\frac{{\lambda }^2}{2}-5\sqrt{2}\lambda -3\sqrt{2}\lambda -16<0$

$\Rightarrow {\lambda }^2-\lambda \sqrt{2}(<0$

$\Rightarrow (\lambda -(4\sqrt{2}-\sqrt{14}\left)\right)(\lambda -(4\sqrt{2}+\sqrt{14}\left)\right)<0$

$\Rightarrow \lambda \in (4\sqrt{2}-\sqrt{14},4\sqrt{2}+\sqrt{14})\to \left(2\right)$

$\because \lambda$ must satisfy both $\left(1\right))$ and $\left(2\right)$

$\Rightarrow \lambda \in (4\sqrt{2}-\sqrt{14},5\sqrt{2})$

$\therefore B)$ Answer.