Circle 1 given x2+y2=4 Circle 2 is given, x2+y2−6x−6y+14=0 Centre (3,3) radius =2 Let P(x1,y1) be a point on circle 1 . ∵ Equation of chord of contact AB is ‌xx1+yy1−3(x+x1)−3(y+y1)+14=0 ‌(x1−3)x+(y1−3)y−3x1−3y1+14=0 Now, equation of circle passing through P,A and B is x2+y2−6x−6y+14+λ[(x1−3)x+(y1−3)y−3x1−3y1+14]=0 We have this circle passes through P(x1,y1)
⇒x12+y12−6x1−6y1+14+λ[(x1−3)x1. +(y1−3)y1−3x1−3y1+14]=0 ⇒x12+y12−6x1−6y1+14+λ[x12+y12. −3x1−3y1−3x1−3y1+14]=x Put x12+y12=4 ‌‌‌⇒4−6x1−6y1+14 ‌‌‌+λ(4−6x1−6y1+14)=0 ‌⇒18−6x1−6y1+14+λ(18−6x1. ‌‌‌−6y1)=0 ‌∵1+λ=0 ‌λ=−1 ⇒ Equation of circle be ‌(x2+y2−6x−6y+14) ‌−[(x1−3)x+(y1−3)y. ‌−3x1−3y1+14]=0 ‌⇒x2+y2−(3+x1)x−(3+y1)y ‌+3x1+3y1=0 ‌‌ Centre ‌(‌