Question

Find the equation of the tangents to circle $x^2+y^2=4$ which make an angle of ${60}^o$ with the positive x-axis in anticlockwise direction.

Answer 1

Given equation of circle $\Rightarrow x^2+y^2=4$

We have to find tangents which

makes an angle ${60}^o$ with $x-axis$

i.e, slope of tangent $=\tan \theta =\tan 60=\sqrt{3}$

let line be $y=mx+cm\to slope$

$y=\sqrt{3}x+C$

$=\sqrt{3}x-y+C=0$

As it si a tangent its perpendicular distance from centre is radius

$\sqrt{3}x-y+C\perp$ar distance to $(0,0)=r$

Here, radius $=2$.

$\therefore \frac{0-0+c}{\sqrt{(\sqrt{3}{)}^2+(1{)}^2}}=2\Rightarrow \frac{0-0+C}{\sqrt{4}}=2$

$\Rightarrow C=2\sqrt{4}$

$C=\pm 4$

$\therefore y=\sqrt{3}x+C$

$\therefore y=\sqrt{3}x\pm 4$