We know that, (1,0),(0,1),(0,0) lies on X−axis, Y−axis and on the origin respectively.
So, (2k,3k) will lie on circle whose diameter is AB.

Equation of the circle will be given by,
⇒(x−1)(x)+(y−1)(y)=0
⇒x2+y2−x−y=0...(i)
Using (2k,3k) in equation (i),
⇒(2k)2+(3k)2−2k−3k=0
⇒13k2−5k=0
⇒k=0,k=135
Hence, k=135.