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

We know that, $(1,0), (0,1), (0,0)$ lies on $X-$axis, $Y-$axis and on the origin respectively.

So, $(2\mathrm{k},3\mathrm{k})$ will lie on circle whose diameter is $\mathrm{AB}$.

Equation of the circle will be given by,

$\Rightarrow (\mathrm{x}-1)\left(\mathrm{x}\right)+(\mathrm{y}-1)\left(\mathrm{y}\right)=0$

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}-\mathrm{y}=0 ...\left(i\right)$

Using $(2\mathrm{k}, 3\mathrm{k})$ in equation $\left(i\right)$,

$\Rightarrow (2\mathrm{k}{)}^2+(3\mathrm{k}{)}^2-2\mathrm{k}-3\mathrm{k}=0$

$\Rightarrow 13{\mathrm{k}}^2-5\mathrm{k}=0$

$\Rightarrow \mathrm{k}=0, \mathrm{k}=\frac{5}{13}$

Hence, $\mathrm{k}=\frac{5}{13}$.