Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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)

best_answer

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

Posted by

prateek

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2025 registration link active at bitsadmission.com; exam schedule, eligibility criteria

anam-khan

BITSAT 2025 registration begins today for 10 BE, 6 MSc, BPharm courses; eligibility, fee

anam-khan

BITSAT 2025 registration from tomorrow at bitsadmission.com; eligibility criteria

Latest Question