To solve the problem, we need to determine the value of \( k \) such that the points \( (2k, 3k) \), \( (1, 0) \), \( (0, 1) \), and \( (0, 0) \) all lie on the same circle.
### Step-by-step Solution:
1. **Identify the Points**:
The points given are:
- \( A(2k, 3k) \)
- \( B(1, 0) \)
- \( C(0, 1) \)
- \( D(0, 0) \)
2. **Determine the Diameter**:
The points \( B(1, 0) \) and \( C(0, 1) \) can be considered as endpoints of a diameter of the circle. The midpoint of \( BC \) will serve as the center of the circle.
\[
\text{Midpoint} = \left( \frac{1 + 0}{2}, \frac{0 + 1}{2} \right) = \left( \frac{1}{2}, \frac{1}{2} \right)
\]
3. **Equation of the Circle**:
The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Here, the center is \( \left( \frac{1}{2}, \frac{1}{2} \right) \).
4. **Finding the Radius**:
The radius can be calculated as the distance from the center to one of the points, say \( B(1, 0) \):
\[
r = \sqrt{\left(1 - \frac{1}{2}\right)^2 + \left(0 - \frac{1}{2}\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}
\]
5. **Circle Equation**:
Plugging the center and radius into the circle equation:
\[
\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{\sqrt{2}}\right)^2
\]
Simplifying gives:
\[
\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{2}
\]
6. **Substituting Point \( A(2k, 3k) \)**:
Since \( A(2k, 3k) \) lies on the circle, we substitute \( x = 2k \) and \( y = 3k \) into the circle equation:
\[
\left(2k - \frac{1}{2}\right)^2 + \left(3k - \frac{1}{2}\right)^2 = \frac{1}{2}
\]
7. **Expanding and Simplifying**:
Expanding both terms:
\[
\left(2k - \frac{1}{2}\right)^2 = 4k^2 - 2k + \frac{1}{4}
\]
\[
\left(3k - \frac{1}{2}\right)^2 = 9k^2 - 3k + \frac{1}{4}
\]
Adding these together:
\[
4k^2 - 2k + \frac{1}{4} + 9k^2 - 3k + \frac{1}{4} = \frac{1}{2}
\]
Simplifying gives:
\[
13k^2 - 5k + \frac{1}{2} = 0
\]
8. **Multiplying through by 2 to eliminate the fraction**:
\[
26k^2 - 10k + 1 = 0
\]
9. **Using the Quadratic Formula**:
Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
k = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 26 \cdot 1}}{2 \cdot 26}
\]
\[
k = \frac{10 \pm \sqrt{100 - 104}}{52} = \frac{10 \pm \sqrt{-4}}{52}
\]
Since the discriminant is negative, there are no real solutions for \( k \).
### Conclusion:
Since the calculations show that the points do not satisfy the circle equation for any real \( k \), we conclude that there is no value of \( k \) for which all four points lie on the same circle.
