Let C1and C2 be the centers 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 PC1QC2is  

Let C1and C2 be the centers 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 PC1QC2is
  

  1. A

    8

  2. B

    4

  3. C

    6

  4. D

    9

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    Solution:

    Let S1x2+y22x2y2=0

    S2x2+y26x6y+14=0 C1(1,1) and C2(3,3)

     

    Radius of S1,PC1=1+1+2=2

    Radius of S2,PC2=9+914=2

     PC1=QC1=PC2=QC2=2

    Now, 2g1g2+2f1f2=2×3+2×3=6+6=12 and 

    c1+c2=142=12

    Here, 2g1g2+2f1f2=c1+c2

    Both circles are orthogonal. So, PC1QC2 is a square.

     Area of PC1QC2=2×2=4 sq. units .

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