The value of k for (2k, 3k), (0, 0), (1, 0) and (0, 1) to be on the circle is

The correct option is 5/13.

To find the value of $k$ such that the points $(2k,3k)$, $(0,0)$, $(1,0)$, and $(0,1)$ lie on the same circle, we can first determine the equation of the circle passing through the three known points: $(0,0)$, $(1,0)$, and $(0,1)$.

The general equation of a circle is given by:
$x^2+y^2+2gx+2fy+c=0$

Step 1: Find the constants by substituting the coordinates of the known points.

1. Since the circle passes through the origin $(0,0)$:
$0^2+0^2+2g\left(0\right)+2f\left(0\right)+c=0\Rightarrow c=0$

2. Since the circle passes through $(1,0)$:
$1^2+0^2+2g\left(1\right)+2f\left(0\right)+0=0$
$1+2g=0\Rightarrow 2g=-1$

3. Since the circle passes through $(0,1)$:
$0^2+1^2+2g\left(0\right)+2f\left(1\right)+0=0$
$1+2f=0\Rightarrow 2f=-1$

Substituting the values of $2g$, $2f$, and $c$ back into the general equation, we obtain the equation of the circle:
$x^2+y^2-x-y=0$

Step 2: Find the value of $k$ using the fourth point.

Since the point $(2k,3k)$ also lies on this circle, its coordinates must satisfy the circle's equation. Let us substitute $x=2k$ and $y=3k$:
${\left(2k\right)}^2+{\left(3k\right)}^2-\left(2k\right)-\left(3k\right)=0$

Simplifying the algebraic equation:
$4k^2+9k^2-5k=0$
$13k^2-5k=0$
$k(13k-5)=0$

This gives us two possible solutions:
$k=0$ or $13k-5=0$

Since $k=0$ represents the origin point $(0,0)$ which is already on the circle, the non-zero value of $k$ is given by:
$13k=5\Rightarrow k=\frac{5}{13}$