Two tangents to the circle $x^2+y^2=4$ at the points $\mathrm{A}$ and $\mathrm{B}$ meet at the point…

Two tangents to the circle $x^2+y^2=4$ at the points $\mathrm{A}$ and $\mathrm{B}$ meet at the point $\mathrm{P}(-4,0)$. Then the area of the quadrilateral $\mathrm{PAOB}, \mathrm{O}$ being the origin, is
  1. $2 \sqrt{3}$ sq. units
  2. $8 \sqrt{3}$ sq. units
  3. $4 \sqrt{3}$ sq. units
  4. $6 \sqrt{3}$ sq. units

Solution

$\begin{aligned} \text { Required area } & =2 \times \text { Area of } \triangle \mathrm{PBO} \\ & =2 \times \frac{1}{2} \times 2 \times 2 \sqrt{3} \\ & =4 \sqrt{3} \text { sq. units }\end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 1)

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