11z1z2 + z3z4 = 0 ----------------(1)
take z1=keiθ...
z1 + z2=0 therefore z2=- keiθ...
Let z3=reiα and z4=Reiβ
Substitute all values we assumed in the eqn --(1)
we get rRcos(α+β)=k2cos2θ &
rRsin(α+β)=k2sin2θ
sqr and add
rR=k2 ----------(2)
α+β=2θ -----------------(3)
Let the eqn passing through the circle be zz + λz + λz + μ =0
Now z1,z2,z3 satisfy this eqn..put the values here..ul get 3 eqns..
Take 2 eqns n solve to get
μ=-k2 --------------(4)
& λeiθ + λe-iθ= 0
Taking conjugate and adding we get
λeiα + λe-iα=-(λeiβ + λe-iβ) ---------------(5)
When we substituted z3 in (1) we got this
r2 + r(λeiα + λe-iα)+μ=0
From this n (5),(4),(2) we get
k4R2 - k2/R(λeiβ + λe-iβ) - k2
= 0
Solving and putting μ=-k2
we get
z4z4 + λz4+λz4 + μ = 0
Thus z4 also satisfies the eqn of the circle.
Hence PROVED.
