Question

The range of values of m for which the line y = mx + 2 cuts the circle $x^2+y^2=1$ at two distinct points, is .

• A. $m\in (-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$
• B. $m\in (-\infty ,-\sqrt{3})$
• C. $m\in (\sqrt{3},\infty )$
• D. $m\in (-\sqrt{3},\sqrt{3})$

Answer 1

The correct option is A $m\in (-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$

$\mathrm{Line}\mathrm{is}:y=\mathrm{mx}+2\mathrm{Circle}\mathrm{is}:x^2+y^2=1\mathrm{On}\mathrm{solving}\mathrm{we}\mathrm{get}:x^2+(\mathrm{mx}+2{)}^2-1=0\Rightarrow x^2+m^2{x}^2+4+4\mathrm{mx}-1=0\Rightarrow (1+m^2)x^2+4\mathrm{mx}+3=0\mathrm{For}\mathrm{two}\mathrm{distinct}\mathrm{points}\mathrm{Discriminant}\mathrm{of}\mathrm{this}\mathrm{equation}\mathrm{must}\mathrm{be}\mathrm{greater}\mathrm{than}0.\Rightarrow D>0\Rightarrow 16m^2-4\cdot (1+m^2)\cdot 3>0\Rightarrow 16m^2-12m^2-12>0\Rightarrow 4m^2-12>0\Rightarrow m^2-3>0\Rightarrow (m-\sqrt{3}\left)\right(m+\sqrt{3})>0\Rightarrow m\in (-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$