Question

If the line $x\cos \theta +y\sin \theta =2$ is the equation of a transverse common tangent to the circles $x^2+y^2=4$ and , $x^2+y^2-6\sqrt{3}x-6y+20=0$ then the value of $\theta$ in degree is :

Answer 1

Given circle are :

$x^2+y^2=4$ centre $=(0,0)$ & radius $=2$ units $\left(r_1\right)$

$x^2+y^2-6\sqrt{3}x-6y+20=0$ centre $=(3\sqrt{3},3)$ & radius $=4$ units $\left(r_2\right)$

Lets find the distance between the centres of two circles $=\sqrt{{\left(3\sqrt{3}\right)}^2+{\left(3\right)}^2}=\sqrt{36}=6$ units $=(2+4)$ units $=r_1+r_2$.

$\therefore$ as the distance between the centres of the two circles is equal to the sum of the radii of the two circles we can say that the cirlces touch each other.

$\therefore$ the equation of the transverse common tangent is given by

$S_1-S_2=0$

Where

$S_1=x^2+y^2-6\sqrt{3}x-6y+20=0$

$S_2=x^2+y^2-4=0$

$\therefore$ $S_1-S_2=0$

$\Rightarrow -6\sqrt{3}x-6y+24=0$

$\Rightarrow \sqrt{3}x+y-4=0$

$\Rightarrow \frac{\sqrt{3}}{2}x+\frac{y}{2}=2$ (Dividing the entire eqn by $2$)

Now comparing with equation given in the question

$x\cos \theta +y\sin \theta =2$

We get $\cos \theta =\frac{\sqrt{3}}{2}\Rightarrow \theta =\frac{\pi }{6}$