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I need help with - Two-dimensional Coordinate Geometry - BITSAT

If the straight line y=mx is outside the circle 

x^{2}+y^{2}-20y+90=0, then

  • Option 1)

    m< 3

  • Option 2)

    |m|<3

  • Option 3)

    m> 3

  • Option 4)

    |m|>3

 
Answers (1)
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As we learnt in 

General form of a circle -

x^{2}+y^{2}+2gx+2fy+c= 0
 

- wherein

centre = \left ( -g,-f \right )

radius = \sqrt{g^{2}+f^{2}-c}

 

 distance of (0, 10) from y=mx

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

& radius =\sqrt{0+100-90}=\sqrt{10}

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

\Rightarrow 1+m^{2}< 10    \Rightarrow m^{2}< 9\Rightarrow -3< m< 3

\Rightarrow \left | m \right |< 3


Option 1)

m< 3

This solution is incorrect

Option 2)

|m|<3

This solution is correct

Option 3)

m> 3

This solution is incorrect

Option 4)

|m|>3

This solution is incorrect

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