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If the straight line $y=m x$ is outside the circle $x^2+y^2-20 y+90=0$, then
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Verified Answer
The correct answer is:
$|m| \lt 3$
If the straight line $y=m x$ is outside the given circle then distance from centre of circle $\gt$ radius of circle
$\frac{10}{\sqrt{1+m^2}}\gt\sqrt{10}$
$\Rightarrow\left(1+m^2\right) \lt 10 \Rightarrow m^2 \lt 9 \Rightarrow|m| \lt 3$.
$\frac{10}{\sqrt{1+m^2}}\gt\sqrt{10}$
$\Rightarrow\left(1+m^2\right) \lt 10 \Rightarrow m^2 \lt 9 \Rightarrow|m| \lt 3$.
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