Four distinct points ( 2 k , 3 k ) , ( 1 , 0 ) , ( 0 , 1 ) and ( 0 , 0 ) lie on a circle for k equal to :

Step-by-step solution

We know that, ( 1 , 0 ) , ( 0 , 1 ) , ( 0 , 0 ) lies on X - axis, Y - axis and on the origin respectively. So, ( 2 k , 3 k ) will lie on circle whose diameter is AB . Equation of the circle will be given by, ⇒ ( x - 1 ) ( x ) + ( y - 1 ) ( y ) = 0 ⇒ x 2 + y 2 - x - y = 0 . . . i Using ( 2 k , 3 k ) in equation i , ⇒ ( 2 k ) 2 + ( 3 k ) 2 - 2 k - 3 k = 0 ⇒ 13 k 2 - 5 k = 0 ⇒ k = 0 , k = 5 13 Hence, k = 5 13 .