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 the quadrilateral PAOB, where O is the origin is

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

Answer 1

The correct option is C $4\sqrt{3}$

Since AP and BP are tangents

$AP=BP$

And triangle $PAO$ is congruent to triangle $PBO$

$\text{Area of PAOB}=2\times \text{Area of triangle APO}$

Given circle is $x^2+y^2=4$

$O=(0,0),r=2=AO$

$P=(-4,0)$

WKT $\angle PAO=90$

$A{P}^2–P{O}^2–A{O}^2=16-4$

$AP=\sqrt{12}$

$Area=\frac{1}{2}\times AP\times AO=\frac{1}{2}\times \sqrt{12}\times 2=\sqrt{12}$

Area of PAOB = $2\sqrt{12}=4\sqrt{3}$