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

Answer

4

Explanation

Because `2g_(1)g_(2)+2f_(1)f_(2)=c_1+c_2`
`2(-1)(-3)+2(-1)(-3)=-2+14rArr` Circles are orthogonal
`r_(1)=sqrt(1+1+2)=2`
`r_(2)=sqrt(9+9-14)=2`
Area `=(1)/(2)xx2xx2+(1)/(2)xx2xx2=4`