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Question

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
12c(g2+f2c
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B
c(g2+f2c)
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C
c(g2+f2c)
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D
g2+f2cc
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Solution

The correct option is C c(g2+f2c)
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,

2 (areaoftriangleOAC)=2×(1/2)OA×AC

=0+0+2g×0+2f×0+cg2+f2c
=c(g2+f2c)

243534_196509_ans_46d06d7145524885bb47dd01099594db.png

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