The points \((1,0),(0,1),(0,0)\) and \((2 \mathrm{k}, 3 \mathrm{k}), \mathrm{k} \neq 0\) are concyclic, if \(\mathrm{k}\) is

The equation of the circle which passes through the points \((1,0),(0,1)\) and \((0,0)\) is
\[
x^{2}+y^{2}-x-y=0 \quad \text{...(i)}
\]
Given that, the point \((2 \mathrm{k}, 3 \mathrm{k})\) is on the circle and form concyclic circle. Then, it satisfies the Eq. (i)
\(\begin{array}{lr} & (2 \mathrm{k})^{2}+(3 \mathrm{k})^{2}-(2 \mathrm{k})-(3 \mathrm{k})=0 \\ \Rightarrow & 4 \mathrm{k}^{2}+9 \mathrm{k}^{2}-5 \mathrm{k}=0 \\ \Rightarrow & 13 \mathrm{k}^{2}-5 \mathrm{k}=0 \\ \Rightarrow & \mathrm{k}(13 \mathrm{k}-5)=0 \\ & \Rightarrow \\ \text { Hence, } & \mathrm{k}=0 \text { or } \mathrm{k}=\frac{5}{13} \\ & \mathrm{k}=\frac{5}{13}\end{array}\)