Q.

AB is a chord of the circle x2+y2=25. The tangents to the circle at A and B intersect  at C . If (2,3) is the midpoint of AB, then the area of quadrilateral OACB (Where O is  origin) is

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a

503

b

50313

c

503

d

5013

answer is B.

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Detailed Solution

Here OB=OA=5 ( radius of circle) OP=(2−0)2+(3−0)2=13 Let ∠OBP=θ, then ∠PCB=θ From ΔOPB,cos⁡(90−θ)=OPOB=135⇒sin⁡θ=135⇒cot⁡θ=2313 Area of quadrilateral OACB=2×12×OB×BC From ΔOBC,cot⁡θ=BCOB=BC5∴ Area of quadrilateral OACB=5×5cot⁡θ=25×2313=50313 sq.units
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