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  • If OA and OB are the tangents from the origin to the circle <img alt="\mathrm{x^2+y^2+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5

If OA and OB are the tangents from the origin to the circle \mathrm{x^2+y^2+2 g x+2 f y+c=0} and C is the centre of the circle, then the area of the quadrilateral OACB 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{\left(g^2+f^2-c\right)}}


Option: 4

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


Answers (1)

best_answer

Since \mathrm{O A=O B} and \mathrm{C A=C B}, so the diagonal OC divides the quadrilateral OACB in two equal right angled triangles, OAC and OBC as shown in figure.

Therefore, the area of the quadrilateral OACB

\mathrm{ \begin{aligned} & =2(\text { area of triangle } O A C)=2 \times(1 / 2) O A \times A C \\\\ & =\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)} \end{aligned} }

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Pankaj

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