Question
Two tangents to the circle $${x^2} + {y^2} = 4$$ at the points $$A$$ and $$B$$ meet at $$P\left( { - 4,\,0} \right).$$ The area of the quadrilateral $$PAOB,$$ where $$O$$ is the origin, is :
A.
$$4$$
B.
$$6\sqrt 2 $$
C.
$$4\sqrt 3 $$
D.
none of these
Answer :
$$4\sqrt 3 $$
Solution :
$$\eqalign{ & {\text{Clearly, }}\sin \,\theta = \frac{2}{4} = \frac{1}{2}\,\,\,\,\,\,\,\,\therefore \theta = {30^ \circ } \cr & {\text{So, ar}}\left( {\Delta POA} \right) = \frac{1}{2}.2.4.\sin \,{60^ \circ } \cr & \therefore \,\,{\text{ar}}\left( {{\text{quad}}{\text{.}}\,PAOB} \right) = 2.\frac{1}{2}.2.4\,\sin \,{60^ \circ } = 8.\frac{{\sqrt 3 }}{2} = 4\sqrt 3 \cr} $$
$$\eqalign{ & {\text{Clearly, }}\sin \,\theta = \frac{2}{4} = \frac{1}{2}\,\,\,\,\,\,\,\,\therefore \theta = {30^ \circ } \cr & {\text{So, ar}}\left( {\Delta POA} \right) = \frac{1}{2}.2.4.\sin \,{60^ \circ } \cr & \therefore \,\,{\text{ar}}\left( {{\text{quad}}{\text{.}}\,PAOB} \right) = 2.\frac{1}{2}.2.4\,\sin \,{60^ \circ } = 8.\frac{{\sqrt 3 }}{2} = 4\sqrt 3 \cr} $$