If the line x cos theta + y sin theta = 2 is | vedclass

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(D) For the circle $x^2 + y^2 = 4$,center $C_1 = (0, 0)$ and radius $r_1 = 2$.For the circle $x^2 + y^2 - 6 \sqrt{3} x - 6y + 20 = 0$,center $C_2 = (3

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$\pi / 6$

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(D) For the circle $x^2 + y^2 = 4$,center $C_1 = (0, 0)$ and radius $r_1 = 2$.For the circle $x^2 + y^2 - 6 \sqrt{3} x - 6y + 20 = 0$,center $C_2 = (3 \sqrt{3}, 3)$ and radius $r_2 = \sqrt{(3 \sqrt{3})^2 + 3^2 - 20} = \sqrt{27 + 9 - 20} = \sqrt{16} = 4$.The distance between centers $C_1C_2 = \sqrt{(3 \sqrt{3} - 0)^2 + (3 - 0)^2} = \sqrt{27 + 9} = 6$.Since $C_1C_2 = r_1 + r_2 = 2 + 4 = 6$,the circles touch each other externally.The common tangent at the point of contact is the radical axis,given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 4) - (x^2 + y^2 - 6 \sqrt{3} x - 6y + 20) = 0$ $\Rightarrow 6 \sqrt{3} x + 6y - 24 = 0$ $\Rightarrow \sqrt{3} x + y = 4$.Dividing by $2$,we get $\frac{\sqrt{3}}{2} x + \frac{1}{2} y = 2$.Comparing this with $x \cos 	heta + y \sin 	heta = 2$,we have $\cos 	heta = \frac{\sqrt{3}}{2}$ and $\sin 	heta = \frac{1}{2}$.Thus,$	heta = \frac{\pi}{6}$.

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 $\theta$ is:

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