Question

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

• A. $\frac{54}{5}sq.unit$
• B. $\frac{1}{3}\sqrt{5}sq.unit$
• C. $\frac{10}{9}\sqrt{5}sq.unit$
• D. $\frac{20}{9}\sqrt{5}sq.unit$

Answer 1

The correct option is A $\frac{54}{5}sq.unit$

REF.Image.

Eqn of circle, $x^2+y^2=4=(r{)}^2\Rightarrow r=2$

$\because PA=PB=\sqrt{(3{)}^2-4}=\sqrt{9-4}=\sqrt{5}$

$OP=\sqrt{3^2-2^2}=\sqrt{9-4}$

$=\sqrt{5}$

$\angle OPA=\angle OPB=\alpha$

In $△PAO,tan\alpha =\frac{2}{3},sin\alpha \frac{2}{\sqrt{5}}$

$cos\alpha =\frac{PM}{3}=\frac{3}{\sqrt{5}}$

$\Rightarrow PM=\frac{9}{\sqrt{5}}$

$sin\alpha =\frac{AM}{3}=\frac{2}{\sqrt{5}}\Rightarrow AM=\frac{6}{\sqrt{5}}$

Area of $△APB,=\frac{1}{2}\times AB\times PM$

$=\frac{1}{2}(2\times AM)\times PM$

$=\frac{1}{2}\times 12\times \frac{6}{\sqrt{5}}\times \frac{9}{\sqrt{5}}$

$\Rightarrow \frac{54}{4}sq.units$