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Let C 1 and C 2 be the centres of the circles x 2 + y 2 – 2x – 2y – 2 = 0 and x 2 + y 2 – 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 PC 1 QC 2 is:

1. 4
2. 6
3. 9
4. 8

Correct answer: 4

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