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If  \mathrm{O A \: and \: O B} are the tangents from the origin to the circle \mathrm{x^2+y^2+2 g x+2 f y+c=0} and \mathrm{C} is the centre of the circle, the area of the quadrilateral \mathrm{O A C B} is
 

Option: 1

\mathrm{\frac{1}{2} \sqrt{c\left(g^2+f^2-c\right)}}

 


Option: 2

\mathrm{\sqrt{c\left(g^2+f^2-c\right)}}
 


Option: 3

\mathrm{c \sqrt{g^2+f^2-c}}
 


Option: 4

\mathrm{\frac{\sqrt{g^2+f^2-c}}{c}}


Answers (1)

best_answer

Since \mathrm{O A=O B \: and \: C A=C B}

The diagonal \mathrm{O C} divides the quadrilateral \mathrm{O A C B} in two equal right angled triangles,

\mathrm{O A C} and \mathrm{O B C.} Given in the figure;

\therefore the area of the quadrilateral \mathrm{O A C B} is

\mathrm{ 2 \text { (area of triangle } O A C)=\quad 2 \times\left(\frac{1}{2}\right) O A \times A C }

\mathrm{=\sqrt{0+0+2 g \times 0+2 f \times 0+c} \sqrt{g^2+f^2-c}=\sqrt{c\left(g^2+f^2-c\right)}}

Hence option 2 is correct.



 

Posted by

Sanket Gandhi

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