From question, the equations of the circles are: x2+y2−2x−2y−2=0 x2+y2−6x−6y+14=0 The two circles are intersected orthogonally if 2g1g2+2f1f2=c1+c2. The above equations are intersecting each other orthogonally, because ⇒2(1)(3)+2(1)(3)=14−2
So, the area of quadrilateral pC1QC2 is given by the formula: ⇒A=2×ar(∆PC1C2) ⇒A=2×(