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

Answer 1

The correct option is C

$4\sqrt{3}$ sq. units

The explanation for the correct answer.

Step 1. Find the radius of the circle

We know that the equation of the circle ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}{\mathit{r}}^{\mathbf{2}}$, $r$ is the radius of the circle.

The equation of a circle is $x^2+y^2=4$

So the radius of the circle is $2$

Step 2. Find the length of $AD$

$∆PAO$ is a right-angle triangle right angle at $A$

$OP=4unit,OA=2unit$

Let $\angle POA=\theta$

$\therefore$$\cos \left(\theta \right)=\frac{2}{4}$

$=\frac{1}{2}$

$\therefore \theta =\frac{\pi }{3}$

$\angle POA=\angle POD=\theta$

$∆ADO$ is a right aright-angle right angle at $D$

$$\begin{gathered} \therefore \sin \left(\frac{\pi }{3}\right)=\frac{AD}{AO} \\ \Rightarrow AD=\frac{\sqrt{3}}{2}·2 \\ \Rightarrow AD=\sqrt{3} \end{gathered}$$

Step 3. Find the area of the quadrilateral

Area of Quadrilateral $PAOB=2$(Area of $∆PAO$)

Area of $∆PAO=\frac{1}{2}\times OP\times AD$

$$\begin{gathered} =\frac{1}{2}\times 4\times \sqrt{3} \\ =2\sqrt{3} \end{gathered}$$

So the area of the quadrilateral is $2\times 2\sqrt{3}=4\sqrt{3}$ sq. units

Hence, option(C) is the correct answer.