The coordinates of the centre and radius of the circle represented by the equation
`(3-2lambda)x^(2)+lambda y^(2)-4x+2y-4=0` are

Options

A. (2, 1), 3
B. (-2, 1), 3
C. (2, 1), 3
D. (2, -1), 1

Explanation

To find the coordinates of the center and the radius of the circle represented by the equation

\[ (3 - 2\lambda)x^2 + \lambda y^2 - 4x + 2y - 4 = 0, \]

we will follow these steps:

### Step 1: Identify the coefficients
The equation can be rewritten in the standard form of a circle. For that, we need to ensure that the coefficients of \(x^2\) and \(y^2\) are equal.

The coefficient of \(x^2\) is \(3 - 2\lambda\) and the coefficient of \(y^2\) is \(\lambda\). For the equation to represent a circle, we set these coefficients equal:

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

### Step 2: Solve for \(\lambda\)
Rearranging the equation gives:

\[ 3 = 3\lambda \implies \lambda = 1. \]

### Step 3: Substitute \(\lambda\) back into the equation
Now, substitute \(\lambda = 1\) back into the original equation:

\[ (3 - 2 \cdot 1)x^2 + 1 \cdot y^2 - 4x + 2y - 4 = 0, \]
which simplifies to:

\[ (3 - 2)x^2 + y^2 - 4x + 2y - 4 = 0 \implies x^2 + y^2 - 4x + 2y - 4 = 0. \]

### Step 4: Rearrange the equation
Next, we rearrange the equation to group the \(x\) and \(y\) terms:

\[ x^2 - 4x + y^2 + 2y = 4. \]

### Step 5: Complete the square
Now, we complete the square for the \(x\) and \(y\) terms.

For \(x^2 - 4x\):
\[ x^2 - 4x = (x - 2)^2 - 4. \]

For \(y^2 + 2y\):
\[ y^2 + 2y = (y + 1)^2 - 1. \]

Substituting these back into the equation gives:

\[ (x - 2)^2 - 4 + (y + 1)^2 - 1 = 4. \]

### Step 6: Simplify the equation
Now, simplifying this:

\[ (x - 2)^2 + (y + 1)^2 - 5 = 4 \implies (x - 2)^2 + (y + 1)^2 = 9. \]

### Step 7: Identify the center and radius
This is now in the standard form of a circle:

\[ (x - h)^2 + (y - k)^2 = r^2, \]

where \((h, k)\) is the center and \(r\) is the radius. From our equation:

- The center \((h, k) = (2, -1)\).
- The radius \(r = \sqrt{9} = 3\).

### Final Answer
Thus, the coordinates of the center are \((2, -1)\) and the radius is \(3\).

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