Let C1 and C2 be the centres of the circles x2+y2−2x−2y−2=0 and x2+y2−6x−6y+14=0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC1QC2 is :
A
4
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B
6
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C
8
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D
9
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Solution
The correct option is A4 Given circle are x2+y2−2x−2y−2=0 and x2+y2−6x−6y+14=0 C1=(1,1),r1=2C2=(3,3),r2=2
As P lies on the circle, so PC1=2,PC2=2C1C2=√4+4=2√2⇒(PC1)2+(PC2)2=(C1C2)2
∴△C1PC2 is a right angle triangle
So, the area of quadrilateral PC1QC2 =2(12×PC1×PC2)=2(12×2×2)=4sq. units