If the equation `(4 lambda - 3) x^(2) + lambda y ^(2) + 6x - 2y + 2 = 0 ` represents a circle, then its centre is

Options

A. (a) (3,-1)
B. (b) (3,1)
C. (c) (-3,1)
D. (d) (-3,-1)

Answer

To determine the center of the circle represented by the equation

\[ (4\lambda - 3)x^2 + \lambda y^2 + 6x - 2y + 2 = 0, \]

we need to rewrite this equation in the standard form of a circle, which is

\[ x^2 + y^2 + 2gx + 2fy + c = 0. \]

### Step 1: Identify the coefficients
We start by identifying the coefficients of \(x^2\) and \(y^2\) in the given equation. The equation can be rewritten as:

\[ (4\lambda - 3)x^2 + \lambda y^2 + 6x - 2y + 2 = 0. \]

### Step 2: Set conditions for a circle
For the equation to represent a circle, the coefficients of \(x^2\) and \(y^2\) must be equal. Thus, we set:

\[ 4\lambda - 3 = \lambda. \]

### Step 3: Solve for \(\lambda\)
Now, we solve for \(\lambda\):

\[ 4\lambda - \lambda - 3 = 0 \implies 3\lambda - 3 = 0 \implies 3\lambda = 3 \implies \lambda = 1. \]

### Step 4: Substitute \(\lambda\) back into the equation
Now, substituting \(\lambda = 1\) back into the original equation gives us:

\[ (4(1) - 3)x^2 + (1)y^2 + 6x - 2y + 2 = 0 \implies (4 - 3)x^2 + y^2 + 6x - 2y + 2 = 0 \implies x^2 + y^2 + 6x - 2y + 2 = 0. \]

### Step 5: Rearranging to standard form
Next, we rearrange this equation into the standard form of a circle:

\[ x^2 + y^2 + 6x - 2y + 2 = 0. \]

### Step 6: Completing the square
We complete the square for \(x\) and \(y\):

1. For \(x\):
\[ x^2 + 6x \rightarrow (x + 3)^2 - 9. \]

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

Substituting these back into the equation gives:

\[ (x + 3)^2 - 9 + (y - 1)^2 - 1 + 2 = 0. \]

This simplifies to:

\[ (x + 3)^2 + (y - 1)^2 - 8 = 0 \implies (x + 3)^2 + (y - 1)^2 = 8. \]

### Step 7: Identify the center
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify the center \((h, k)\):

\[ h = -3, \quad k = 1. \]

Thus, the center of the circle is

\[ (-3, 1). \]

### Final Answer
Therefore, the center of the circle is \((-3, 1)\).

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