Question

From the point $P(4,3)$ tangents $PA$ and $PB$ are drawn to the circle $x^2+y^2=9$ Than area of the triangle $PAB$ is-

• A. $\frac{192}{25}$sq units
• B. $\frac{16}{9}$ sq units
• C. $\frac{21\sqrt{3}}{16}squnits$
• D. $\frac{25\sqrt{3}}{9}squnits$

Answer 1

The correct option is A $\frac{192}{25}$sq units

From the figure in the question,,since $PQ=PR=\sqrt{4^2+3^2-9}=\sqrt{16+9-9}=4$

$\angle CPQ=\angle CPR=\alpha$

In $△PQC,\tan \alpha =\frac{3}{4},\sin \alpha =\frac{3}{5}$

$\cos \alpha =\frac{PM}{4}=\frac{4}{5}$

$\Rightarrow PM=\frac{16}{5}$

$\sin \alpha =\frac{QM}{4}=\frac{3}{5}$

$\Rightarrow QM=\frac{12}{5}$

Area of $△PQR=\frac{1}{2}\times QR\times PM$

$=\frac{1}{2}\times 2QM\times PM$

$=QM\times PM$

$=\frac{12}{5}\times \frac{16}{5}=\frac{192}{25}$