If OA and OB are the tangents from the origin to the circle x2+y2+2gx+2fy+c=0, and C is the centre of the circle, the area of the quadrilateral OACB is
A
12√c(g2+f2−c
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B
√c(g2+f2−c)
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C
c√(g2+f2−c)
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D
√g2+f2−cc
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Solution
The correct option is C√c(g2+f2−c) Since OA=OB and CA=CB
The diagonal OC divides the quadrilateral OACB in two equal right-angled triangles, ΔOAC and ΔOBC.
Therefore, the area of the quadrilateral OACB is given as,