Pbe a point on circle (x−1)2+(y−1)2=1 So, coordinate of P will be of form P(1+cosθ,1+sinθ) Given, A(1,4) and B(1,−5) Then, (PA)2+(PB)2 =cos2θ+(sinθ−3)2+cos2θ+(sinθ+6)2 =2cos2θ+2sin2θ−6sinθ+12sinθ+45=47+6sinθ Now, (PA)2+(PB)2=47+6sinθ is Maximum, when sinθ=1 ⇒θ=
π
2
⇒cosθ=0 ∴P(1,2),A(1,4),B(1,−5) ∴P,B are collinear points on the line x=1