Question

Two tangents to the circle $x ^ { 2 } + y ^ { 2 } = 4$ at the points A and B meet at P (– 4, 0). The area of quadrilateral PAOB, where O is the origin, is

• A. 4
• B. $6 \sqrt { 2 }$
• C. $4 \sqrt { 3 }$
• D. None of these

Answer 1

Answer: (C)

$4 \sqrt { 3 }$

Answer 2

Clearly, $\sin \theta = \frac { 2 } { 4 } = \frac { 1 } { 2 }$ ,

$\therefore \theta = 30 ^ { \circ }$

So area $( \triangle P O A ) = \frac { 1 } { 2 } \cdot 2 \cdot 4 \cdot \sin 60 ^ { \circ }$

$\therefore$Area (quadrilateral PAOB)

Trick : Area of quadrilateral $= r \sqrt { S _ { 1 } } = 2 \sqrt { 12 } = 4 \sqrt { 3 }$