Question

If the line $y-mx+m-1=0$ cuts the circle $x^2+y^2-4x-4y+4=0$ at two real points, then m belongs to

• A. $[1,1]$
• B. $[-2,2]$
• C. $(-\infty ,\infty )$
• D. $[-4,4]$

Answer 1

The correct option is C $(-\infty ,\infty )$

Intersection point of line and circle

$x^2+{[mx-(m-1\left)\right]}^2-4x-4[mx-(m-1\left)\right]+4=0(1+m^2)x^2-[4+4m+2m(m-1\left)\right]x+{(m-1)}^2+4(m-1)+4=0(1+m^2)x^2-[2m^2+2m+4]x+{(m-1+2)}^2=0(1+m^2)x^2-[2m^2+2m+4]x+{(m+1)}^2=0$

For $x$ to be real $D\ge 0$

${(2m^2+2m+4)}^2-4(1+m^2){(m+1)}^2\ge 0{\left(2m\right(m+1)+4)}^2-4(1+m^2){(m+1)}^2\ge 04m^2{(m+1)}^2+16+16m(m+1)-4(1+m^2){(m+1)}^2\ge 04m^2{(m+1)}^2+16+16m(m+1)-4{(m+1)}^2-4m^2{(m+1)}^2\ge 016+16m(m+1)-4{(m+1)}^2\ge 016+(m+1)[16m-4m-4]\ge 016+(m+1)[12m-4]\ge 016+4(m+1)(3m-1)\ge 04+(m+1)(3m-1)\ge 0(3m^2+2m+3)\ge 0$

Discriminant $=4-4\times 3\times 3=-32$

For all real values of $m$, the coordinates of $x$ will be real $m\in (-\infty ,\infty )$