The line xcosA +ysinA =p intersects the circle x²+y²=4 at A and B . If the chord AB makes an angle 30^o at a point on the circumference of the circle , then A) p²=3 B)p²=4cos215degrees C) p²=2 D)p²=6

The equation of the given circle is :
$x^2 + y^2 = 4$

So ,the centre of the given circle is at the origin O(0, 0) and its radius is 2 units.

Now AB is the chord of the given circle whose equation is,

$x \cos \alpha + y \sin \alpha = p$

Here p is the length of the perpendicular drawn from the origin on the chord AB .

Now, OL = p.



$$\begin{gathered} \mathrm{In} ∆\mathrm{ALO} \mathrm{and} ∆\mathrm{BLO}, \mathrm{we} \mathrm{have}, \\ \angle \mathrm{ALO} = \angle \mathrm{BLO} \left[90°\mathrm{each}\right] \\ \mathrm{OA} = \mathrm{OB} \left[\mathrm{radii} \mathrm{of} \mathrm{same} \mathrm{circle}\right] \\ \mathrm{OL} = \mathrm{OL} \left(\mathrm{common}\right) \\ ∆\mathrm{ALO} \cong ∆\mathrm{BLO} \left[\mathrm{RHS}\right] \\ \Rightarrow \angle \mathrm{AOL} = \angle \mathrm{BOL} \left[\mathrm{CPCT}\right] .......\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \mathrm{we} \mathrm{have}, \\ \angle \mathrm{ACB} = 30° \left(\mathrm{given}\right) \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{angle} \mathrm{subtended} \mathrm{by} \mathrm{a} \mathrm{chord} \mathrm{at} \mathrm{the} \mathrm{centre} \mathrm{is} \mathrm{twice} \mathrm{the} \mathrm{angle} \mathrm{subtended} \\ \mathrm{by} \mathrm{the} \mathrm{same} \mathrm{chord} \mathrm{at} \mathrm{any} \mathrm{point} \mathrm{on} \mathrm{the} \mathrm{circumference} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \Rightarrow \angle \mathrm{AOB} = 60° \\ \mathrm{Now}, \\ \angle \mathrm{AOL} = \angle \mathrm{BOL} = 30° \left[\mathrm{using} \left(1\right)\right] \\ \mathrm{In} ∆\mathrm{OLB}, \\ \cos 30° = \frac{\mathrm{base}}{\mathrm{hypotenuse}} \\ \Rightarrow \frac{\sqrt{3}}{2} = \frac{\mathrm{OL}}{\mathrm{OB}} \\ \Rightarrow \frac{\sqrt{3}}{2} =\frac{\mathrm{p}}{2} \left[\mathrm{as}, \mathrm{radius} = \mathrm{OA} =\mathrm{OB} = 2 \mathrm{units}\right] \\ \Rightarrow p = \sqrt{3} \\ \Rightarrow {\mathit{p}}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{3} \end{gathered}$$

Hence, option (a) is correct.