Here, ∣z−2∣≤1 Put z=x+iy

(x−2)2+y2≤1
Also, z(1+i)+zˉ(1−i)≥4
Gives x−y≥2
Let point on circle be A(2+cosθ,sinθ)
θ∈[−43π,4π]
Let P be (25,0)
So,
∣z−25∣2=(AP)2=(2+cosθ−25)2+sin2θ
=cos2θ−cosθ+41+sin2θ
=45−cosθ
For (AP)2 to be maximum θ=−43π
(AP)2=45+21=4252+4=45+22