Question

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

• A. $m<3$
• B. $\left|m\right|<3$
• C. $m>\frac{1}{3}$
• D. $\left|m\right|>3$

Answer 1

The correct option is C $m>\frac{1}{3}$

Given circle, $x^2+y^2-20y+90=0$

Center, $(-g,-f)(10,0)$

$d\to$distance between center and line $y=mx$

$d=\frac{10m}{\sqrt{m^2+1}}\to 1$ $r$=radius; $r=\sqrt{10}$

As the line is outside the circle then $d>r$

$\frac{10m}{\sqrt{m^2+1}}>\sqrt{10}$

Squaring on both sides

$\frac{100m^2}{1+m^2}>10$

$100m^2>10+10m^2$

$90m^2>10$

$m^2>1/9\Rightarrow m>1/3$

$\therefore m>\frac{1}{3}$