Question

If straight line $y=mx$ is outside the circle $x^2+y^2-20y+90=0$, then

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

Answer 1

The correct option is D $|m|>3$

$x^2+y^2-20y+90=0⟶\left(1\right)$

Straight line is $y=mx$ putting $y=mx$ in $\left(1\right)$, we get

$x^2+m^2{x}^2-20mx+90=0$

$x^2(1+m^2)-20mx+90=0$

$\Rightarrow D>0$ (Since the line lies outside the circle.)

${400m}^2-4\times 90(1+m^2)>0$

$\Rightarrow {400m}^2-360(1+m^2)>0$

$\Rightarrow {400m}^2-360-360m^2>0$

$\Rightarrow {40m}^2>360$

$\Rightarrow m^2>9$

$\Rightarrow \left|m\right|>3$