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  be the centers of the circles  and  respectively. If   are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $P{C}_{1}Q{C}_{2}$is

1. A

8

2. B

4

3. C

6

4. D

9

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### Solution:

Let ${S}_{1}\equiv {x}^{2}+{y}^{2}-2x-2y-2=0$

Radius of ${S}_{1},P{C}_{1}=\sqrt{1+1+2}=2$

Radius of ${S}_{2},P{C}_{2}=\sqrt{9+9-14}=2$

Now, $2{g}_{1}{g}_{2}+2{f}_{1}{f}_{2}=2×3+2×3=6+6=12$ and

${c}_{1}+{c}_{2}=14-2=12$

Here, $2{g}_{1}{g}_{2}+2{f}_{1}{f}_{2}={c}_{1}+{c}_{2}$

$\therefore$Both circles are orthogonal. So, $P{C}_{1}Q{C}_{2}$ is a square.

$\therefore$ Area of .

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