question_answer

Any circle through the point of intersection of the lines \[x+\sqrt{3}y=1\] and \[\sqrt{3}x-y=2\]if intersects these lines at points P and Q, then the angle subtended by the arc PQ at its centre is                                                                    [MP PET 1998]

A)            \[{{180}^{o}}\]

B)            \[{{90}^{o}}\]

C)            \[{{120}^{o}}\]                    

D)            Depends on centre and radius

Correct Answer: A

Solution :

           Let the point of intersection of two lines is A.                    \[\therefore \] The angle subtended by PQ on centre C                    \[=\]Two times the angle subtended by PQ on point A.                    For \[x+\sqrt{3}y=1\], \[{{m}_{1}}=\frac{-1}{\sqrt{3}}\] and For \[\sqrt{3}x-y=2,\] \[{{m}_{2}}=\sqrt{3}\]                    \[\because \]\[{{m}_{1}}\times {{m}_{2}}=\frac{-1}{\sqrt{3}}\times \sqrt{3}=-1\], \[\therefore \ \angle A={{90}^{o}}\]                    \[\therefore \]The angle subtended by arc PQ at its centre \[=2\times {{90}^{o}}={{180}^{o}}\]                    Trick: Given lines are perpendicular to each other, so PQ passes through centre of circle, hence arc makes \[{{180}^{o}}\] to centre.