The points (1,0),(0,1),(0,0) and (2 k , 3 k ), k 0 are concyclic, if k is

Step-by-step solution

The equation of the circle which passes through the points (1,0),(0,1) and (0,0) is x²+y²-x-y=0 ...(i) Given that, the point (2 k , 3 k ) is on the circle and form concyclic circle. Then, it satisfies the Eq. (i) array lr & (2 k )²+(3 k )²-(2 k )-(3 k )=0 & 4 k ²+9 k ²-5 k =0 & 13 k ²-5 k =0 & k (13 k -5)=0 & Hence, & k =0 or k = 5 13 & k = 5 13 array