Question

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

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

Answer 1

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

Length of tangent from $P$ to the given circle is
$L=\sqrt{S_1}=\sqrt{5}$ units
So required area of triangle is $=\frac{L^{3}r}{L^2+r^2}$
$=\frac{10\sqrt{5}}{9}$ sq. units

Alternate Solution:
Equation of $AB$ is, $T=0$, i.e, $3x+y\cdot 0=4$
$\Rightarrow x=\frac{4}{3},OD=\frac{4}{3}$.
Hence $DP=3-4/3=5/3$
In right angled triangle $△ADO$,
$A{D}^2=O{A}^2-O{D}^2$
$\Rightarrow A{D}^2=4-\frac{16}{9}$
$\Rightarrow AD=\frac{2\sqrt{5}}{3}$
$\Rightarrow AB=\frac{4\sqrt{5}}{3}$
$\Rightarrow$ Area of $△PAB=\frac{1}{2}\cdot \frac{4\sqrt{5}}{3}\cdot \frac{5}{3}$
$=\frac{10}{9}\sqrt{5}$ sq unit