The equation
`lambda^(2)x^(2)+(lambda^(2)-5 lambda+4)xy+(3lambda-2)y^(2)-8x+12y-4=0` will represent a circle, if `lambda=`

Options

A. 1
B. 4
C. 2
D. none of these

Explanation

To determine the values of \( \lambda \) for which the given equation represents a circle, we need to analyze the equation:

\[ \lambda^2 x^2 + (\lambda^2 - 5\lambda + 4)xy + (3\lambda - 2)y^2 - 8x + 12y - 4 = 0 \]

A conic section represents a circle if the following conditions are satisfied:
1. The coefficient of \( x^2 \) is equal to the coefficient of \( y^2 \).
2. The coefficient of \( xy \) is equal to zero.

### Step 1: Coefficient of \( x^2 \) and \( y^2 \)

The coefficient of \( x^2 \) is \( \lambda^2 \) and the coefficient of \( y^2 \) is \( 3\lambda - 2 \).

Set these coefficients equal to each other:

\[ \lambda^2 = 3\lambda - 2 \]

Rearranging gives us:

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

### Step 2: Solve the quadratic equation

We can factor the quadratic equation:

\[ (\lambda - 1)(\lambda - 2) = 0 \]

Thus, the solutions are:

\[ \lambda = 1 \quad \text{or} \quad \lambda = 2 \]

### Step 3: Coefficient of \( xy \)

Next, we need to set the coefficient of \( xy \) to zero. The coefficient of \( xy \) is \( \lambda^2 - 5\lambda + 4 \).

Set this equal to zero:

\[ \lambda^2 - 5\lambda + 4 = 0 \]

### Step 4: Solve the quadratic equation

Factoring gives us:

\[ (\lambda - 1)(\lambda - 4) = 0 \]

Thus, the solutions are:

\[ \lambda = 1 \quad \text{or} \quad \lambda = 4 \]

### Step 5: Combine the results

Now we have two sets of solutions:
1. From the first condition: \( \lambda = 1 \) or \( \lambda = 2 \)
2. From the second condition: \( \lambda = 1 \) or \( \lambda = 4 \)

The common solution from both conditions is:

\[ \lambda = 1 \]

### Final Answer

Thus, the equation will represent a circle if:

\[ \lambda = 1 \]