Question

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

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

Answer 1

Answer: (B)

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