Conic Sections Question 206
Question: If OA and OB are the tangents form the origin to the circle $ x^{2}+y^{2}+2gx+2fy+c=0 $ and C is the centre of the circle, the area of the quadrilateral OACD is
Options:
A) $ \frac{1}{2}\sqrt{c(g^{2}+f^{2}-c)} $
B) $ \sqrt{c(g^{2}+f^{2}-c)} $
C) $ c\sqrt{g^{2}+f^{2}-c} $
D) $ \frac{\sqrt{g^{2}+f^{2}-c}}{c} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Area of quadrilateral $ =2[areaof\Delta OAC] $
$ =2.\frac{1}{2}OA.AC=\sqrt{S_1}.\sqrt{g^{2}+f^{2}-c} $ Point is $ (0,0)\Rightarrow S_1=c, $
$ \therefore $ Area $ =\sqrt{c(g^{2}+f^{2}-c)} $
BETA
