Question:
let $C_{1}$ and $C_{2}$ be the centres of the circles $x^{2}+y^{2}-2 x-2 y-2=0$ and $x^{2}+y^{2}-6 x-6 y+14=0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $\mathrm{PC}_{1} \mathrm{QC}_{2}$ is :
Correct Option: , 4
Solution:
Area $=2 \times \frac{1}{2} \cdot 4=2$
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