logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

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?

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

Can you answer this question?
 
 

1 Answer

0 votes
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.
answered Feb 11, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...