The correct option is D (52,0)
Given that L1:y=2x
L2:y=−2x
L3:x=1
L1L2+λL2L3+μL3L1=0
⇒(y−2x)(y+2x)+λ(y+2x)(x−1)+μ(y−2x)(x−1)=0
⇒(2λ−2μ−4)x2+y2+(λ+μ)xy+2(μ−λ)x=0
Since, above equation is circle with center ((λ−μ),0)
Therefore, 2λ−2μ−4=1 and λ+μ=0
⇒λ−μ=52
Therefore, circumcentre is (52,0)
Ans: B