x+(k−1)y+3=0 2x+k2y−4=0 ∵ there 2 lines are perpendicular
1
1−k
−(−
2
k2
)=−1 2=k2(1−k) ⇒k3−k2+2=0 ⇒k=−1 ∴Linesare x-2 y+3=0 $ 2x+y−4=0 \} x=1 \;\; y=2$ ∴ Centre is (1, 2) Circle will be x2+y2−2x−4y=0 Line x−y+2=0 will intersect at A(−1,1) and B (2, 4) ∴(AB)2=(2+1)2+(4−1)2 =9+9=18