Question

If the straight line $mx-y=1+2x$ intersects the circle $x^2+y^2=1$ at least at one point, then the set of values of m is

• A. $(-4/3,0)$
• B. $(-4/3,4/3)$
• C. $(0,4/3)$
• D. all of these

Answer 1

The correct option is D all of these

$y=(m–2)x–1$ is the line given and it intersects $x^2+y^2=1$ at least at one point

$⟹x^2+\left(\right(m-2)x-1{)}^2–1=0$ must have real roots

$\Delta \ge 0$

$x^2+(m-2{)}^2{x}^2+1–2(m-2)x=1$

$\left(\right(m-2{)}^2+1\left)x^{-}2\right(m–2)x=0$

$b^2–4ac\ge 0$

$⟹4(m–2{)}^2–4\times ((m–2{)}^2+1)\times 0\ge 0$

$⟹(m-2{)}^2\ge 0$

Which is always true since square of any number is non negative.

All the options satisfy .