Question

If the straight line $y=mx$ lies outside the circle $x^2+y^2-20y+90=0$,then the value of $m$ will satisfy

• A. $m<3$
• B. $\left|m\right|<3$
• C. $m>3$
• D. $\left|m\right|>3$

Answer 1

Answer: (B)

$\left|m\right|<3$

The intersection of line and circle is
$x^2+x^2{m}^2-20mx+90=0$
$\Rightarrow x^2(1+m^2)-20mx+90=0$
Since, the line lies outside the circle.
$\therefore$ $D<0$
$\Rightarrow 400m^2-4\times 90(1+m^2)<0$
$\Rightarrow 40m^2<360\Rightarrow m^2<9$
$\Rightarrow \left|m\right|<3$