Question

If the line $y=mx-(m-1)$ cuts the circle $x^2+y^2=4$ at two real and distinct points, then

• A. $m\in (1,2)$
• B. $m=1$
• C. $m=2$
• D. $m\in R$

Answer 1

The correct option is D $m\in R$

Given that line

$y=mx-(m-1)$ __(1)

cuts the circle

$x^2+y^2=4$ __(2)

put the value of y equation (2) and we get,

$x^2+(mx-(m-1){)}^2=4$

$\Rightarrow x^2+m^2{x}^2+(m-1{)}^2-2mx(m-1)=4$

$\Rightarrow x^2+m^2{x}^2+m^2+1-2m-2m^{2}x+2mx=4$

$\Rightarrow x^2(1+m^2)+x(2m-2m^2)+(m^2-2m)=4-1$

$\Rightarrow (1+m^2)x^2+2mx(1-m)+(m^2-2m-3)=0$

compare that,

$\Rightarrow A{x}^2+Bx+C=0$

now,

$A=(1+m^2)$

$B=2m(1-m)$

$C=(m^2-2m-3)$

Also, given that roots are real and distinct.

$D>0$

$B^2=4AC>0$

$\left[2m\right(1-m){]}^2-4(1+m^2\left)\right(m^2-2m-3)>0$

$\Rightarrow 4m^2(1+m^2-2m]-4[m^2-2m-3+m^4-2m^3-3m^2]>0$

$\Rightarrow 4[m^2+m^4-2m^3-m^2+2m+3-m^4+2m^3+3m^2]>0$

$\Rightarrow 3m^2+2m+3>0$

using quadratic formula

$\Rightarrow m=\frac{-2\pm \sqrt{4-36}}{6}>0$

$m=\frac{-2\pm \sqrt{-32}}{6}>0$

$m=\frac{-2\pm 4i\sqrt{2}}{6}>0$ $\because i=\sqrt{-1}$

$m=\frac{-1\pm 2i\sqrt{2}}{3}>0$

Hence this is the answer