Four distinct points `(2lambda,3lambda)`, (1,0), (0, 1) and (0,0) lie on a circle for

Options

A. all integral values of `lambda`
B. `0 lt lambda lt 1`
C. `lambda lt 0`
D. only one value of `lambda`

Answer

To find the condition on \( \lambda \) such that the points \( (2\lambda, 3\lambda) \), \( (1, 0) \), \( (0, 1) \), and \( (0, 0) \) lie on the same circle, we can follow these steps:

### Step 1: Identify the points
The points given are:
- \( A(1, 0) \)
- \( B(0, 1) \)
- \( C(0, 0) \)
- \( D(2\lambda, 3\lambda) \)

### Step 2: Find the equation of the circle
The points \( A \) and \( B \) can be used to find the equation of the circle. The general equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:

\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]

For our points \( A(1, 0) \) and \( B(0, 1) \):
- \( x_1 = 1, y_1 = 0 \)
- \( x_2 = 0, y_2 = 1 \)

Substituting these values into the equation gives:

\[ (x - 1)(x - 0) + (y - 0)(y - 1) = 0 \]

### Step 3: Simplify the equation
Expanding the equation:

\[ x(x - 1) + y(y - 1) = 0 \]

This simplifies to:

\[ x^2 - x + y^2 - y = 0 \]

Rearranging gives:

\[ x^2 + y^2 - x - y = 0 \]

### Step 4: Substitute the fourth point
Now, we substitute the point \( D(2\lambda, 3\lambda) \) into the circle equation:

\[ (2\lambda)^2 + (3\lambda)^2 - (2\lambda) - (3\lambda) = 0 \]

Calculating each term:

\[ 4\lambda^2 + 9\lambda^2 - 2\lambda - 3\lambda = 0 \]

This simplifies to:

\[ 13\lambda^2 - 5\lambda = 0 \]

### Step 5: Factor the equation
Factoring out \( \lambda \):

\[ \lambda(13\lambda - 5) = 0 \]

### Step 6: Solve for \( \lambda \)
Setting each factor to zero gives:

1. \( \lambda = 0 \)
2. \( 13\lambda - 5 = 0 \Rightarrow \lambda = \frac{5}{13} \)

### Step 7: Determine distinct points
The point \( (2\lambda, 3\lambda) \) must be distinct from the origin \( (0, 0) \). If \( \lambda = 0 \), then \( (2\lambda, 3\lambda) = (0, 0) \), which is not distinct. Therefore, we discard \( \lambda = 0 \).

Thus, the valid solution is:

\[ \lambda = \frac{5}{13} \]

### Conclusion
The condition for \( \lambda \) such that the four points lie on the same circle is:

\[ \lambda = \frac{5}{13} \]