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# Find the area of triangle formed by the tangents (4,3) to the circle $x^2+y^2=9$ and the line segment joining their point of contact?

$\begin{array}{1 1}(a)\;\large\frac{162}{35}\\(b)\;\large\frac{152}{35}\\(c)\;\large\frac{192}{25}\\(d)\;\large\frac{142}{25}\end{array}$

Since $PQ=PR=\sqrt{4^2+3^2-9}=4$
$\angle CPQ=\angle CPR =\alpha$
In $\Delta PQC,\tan \alpha =\large\frac{3}{4}$$,\sin\alpha =\large\frac{3}{5} \cos\alpha=\large\frac{PM}{4}=\frac{4}{5} PM=\large\frac{16}{5} \sin\alpha=\large\frac{QM}{4}=\frac{3}{5} QM=\large\frac{12}{5} Area of \Delta PQR=\large\frac{1}{2}$$QR.PM$
$\Rightarrow \large\frac{1}{2}$$(2QM).PM=QM.PM$
$\Rightarrow \large\frac{12}{5}\times \frac{16}{5}$
$\Rightarrow \large\frac{192}{25}$
Hence (c) is the correct answer.