Question

Tangents $PA$ and $PB$ are drawn to $x^2+y^2=4$ from the point $P(3,0).$ Then the area (in sq. units) of $△PAB$ is

• A. $\frac{5}{9}\sqrt{5}$
• B. $\frac{1}{3}\sqrt{5}$
• C. $\frac{10}{9}\sqrt{5}$
• D. $\frac{20}{3}\sqrt{5}$

Answer 1

The correct option is C $\frac{10}{9}\sqrt{5}$


Length of tangent from $P$ to the given circle is
$L=\sqrt{S_1}=\sqrt{9-4}=\sqrt{5}$ units
Equation of chord of contact for point $(3,0)$ is
$x\left(3\right)+y\left(0\right)=4$
$\Rightarrow 3x-4=0$
Perpendicular distance to the chord of contact from the centre $(0,0)$ of the circle is
$p=\frac{|3\times 0-4|}{\sqrt{3^2}}$
$\Rightarrow p=\frac{4}{3}$

So, required area of triangle
$=ar\left(AOBP\right)-ar\left(OAB\right)$
$=r\sqrt{S_1}-\left(p\sqrt{r^2-p^2}\right)$
$=2\sqrt{5}-\frac{4}{3}\sqrt{4-\frac{16}{9}}$
$=2\sqrt{5}-\frac{4}{3}\sqrt{\frac{20}{9}}$
$=2\sqrt{5}-\frac{8}{9}\sqrt{5}$
$=\frac{10\sqrt{5}}{9}$ sq. units