Question

The value(s) of m for which the line $y=mx$ lies wholly outside the circle $x^2+y^2-2x-4y+1=0$, is(are)

• A. $m\in (-4/3,0)$
• B. $m\in (-4/3,0]$
• C. $m\in (0,4/3)$
• D. none of these

Answer 1

Answer: (A)

$m\in (-4/3,0)$

Given circle $x^2+y^2-2x-4y+1=0$, line $y=mx$

$x^2+y^2-2x-4y+1=0$

$\Rightarrow x^2-2x+1+y^2-4y+4-4=0$

$\Rightarrow (x-1{)}^2+(y-2{)}^2-4=0$

For y = mx to lie completely outside $(x-1{)}^2+(y–2{)}^2=2^2$ then the equation $(x-1{)}^2+(mx-2{)}^2–4=0$ must have imaginary roots

$⟹\Delta <0$

$⟹(m^2+1)x^2–x(2+4m)+1=0$

$b^2–4ac<0$

$⟹(2+4m{)}^2–4(m^2+1\left)\right(1)<0$

$⟹16m^2+4+16m-4m^2-4<0$

$⟹4(3m^2+4m)+0<0$

$⟹3m^2+4m<0$

$⟹m(3m+4)<0$

$⟹m(m+\frac{4}{3})<0$

$⟹\frac{-4}{3}<m<0$

$⟹m\in (\frac{-4}{3},0)$