Question

The range of values of $\alpha$ for which the point $(\alpha ,\alpha )$ lies in the interior part of smaller segment of $x^2+y^2=4$ intercepted by the line $3x+4y+7=0,$ is

• A. $(-\infty ,\sqrt{2})$
• B. $(-\sqrt{2},\sqrt{2})$
• C. $(-\sqrt{2},-1)$
• D. $\varphi$

Answer 1

The correct option is C $(-\sqrt{2},-1)$

Given equation of the circle,
$S:x^2+y^2=4$
As point $(\alpha ,\alpha )$ lies inside the circle, then
$S_1<0\Rightarrow {\alpha }^2+{\alpha }^2-4<0\Rightarrow {\alpha }^2-2<0\Rightarrow \alpha \in (-\sqrt{2},\sqrt{2})\dots \left(1\right)$

Also, centre of circle $(0,0)$ and given point $(\alpha ,\alpha )$ lie on opposite sides of $3x+4y+7=0,$ hence
$(3\cdot 0+4\cdot 0+7)\cdot (3\alpha +4\alpha +7)<0$
$\Rightarrow 7\alpha +7<0\Rightarrow \alpha \in (-\infty ,-1)\dots \left(2\right)$

Hence, final range of $\alpha$ is $\left(1\right)\cap \left(2\right)$
$\Rightarrow \alpha \in (-\sqrt{2},-1)$