`x ^(2) + y^(2) + (2K -1) xy -2x + 4y + 3=0` represents the equation of a circle, find k and radius of the circle ?

Options

A. `-2, sqrt2`
B. `1/2, sqrt2`
C. `-2, sqrt3`
D. `2,sqrt3`

Answer

To solve the equation \( x^2 + y^2 + (2K - 1)xy - 2x + 4y + 3 = 0 \) and determine the values of \( k \) and the radius of the circle, we can follow these steps:

### Step 1: Identify the coefficients
The given equation can be compared with the general second degree equation:
\[ Ax^2 + By^2 + 2Hxy + 2Gx + 2Fy + C = 0 \]
Here, we have:
- \( A = 1 \) (coefficient of \( x^2 \))
- \( B = 1 \) (coefficient of \( y^2 \))
- \( H = 2K - 1 \) (coefficient of \( xy \))
- \( G = -1 \) (coefficient of \( x \))
- \( F = 2 \) (coefficient of \( y \))
- \( C = 3 \) (constant term)

### Step 2: Apply the conditions for a circle
For the equation to represent a circle, the following conditions must be satisfied:
1. \( A = B \)
2. \( H = 0 \)

From our coefficients:
- We already have \( A = B = 1 \), which satisfies the first condition.
- For the second condition, we set \( H = 0 \):
\[ 2K - 1 = 0 \]
Solving for \( K \):
\[ 2K = 1 \implies K = \frac{1}{2} \]

### Step 3: Substitute \( K \) back into the equation
Now we substitute \( K \) back into the original equation:
\[ x^2 + y^2 + (2 \cdot \frac{1}{2} - 1)xy - 2x + 4y + 3 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 2x + 4y + 3 = 0 \]

### Step 4: Rewrite the equation in standard form
To find the center and radius, we need to rewrite the equation in the standard form of a circle:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
We can complete the square for \( x \) and \( y \):
1. For \( x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
2. For \( y \):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

Substituting these back into the equation:
\[ (x - 1)^2 - 1 + (y + 2)^2 - 4 + 3 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y + 2)^2 - 2 = 0 \]
\[ (x - 1)^2 + (y + 2)^2 = 2 \]

### Step 5: Identify the center and radius
From the standard form \( (x - h)^2 + (y - k)^2 = r^2 \):
- The center \( (h, k) = (1, -2) \)
- The radius \( r = \sqrt{2} \)

### Final Answer
Thus, we have:
- \( K = \frac{1}{2} \)
- Radius of the circle \( r = \sqrt{2} \)