`x ^(2) + hxy + y^(2) -6x -2y +k=0 ` is the equation of the circle and 2 is the radius of the circle, then find the values of h and k ?

Options

A. `h =0, k =-6`
B. `h =0, k =6`
C. ` h=-3, k=6`
D. `h =3, k =6`

Answer

To solve the equation \( x^2 + hxy + y^2 - 6x - 2y + k = 0 \) and find the values of \( h \) and \( k \) given that it represents a circle with a radius of 2, we can follow these steps:

### Step 1: Identify the standard form of the circle's equation
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \( g \), \( f \), and \( c \) are constants.

### Step 2: Compare coefficients
From the given equation \( x^2 + hxy + y^2 - 6x - 2y + k = 0 \), we can identify:
- Coefficient of \( x^2 \) is \( 1 \)
- Coefficient of \( y^2 \) is \( 1 \)
- Coefficient of \( xy \) is \( h \)
- Coefficient of \( x \) is \( -6 \) (thus \( 2g = -6 \) or \( g = -3 \))
- Coefficient of \( y \) is \( -2 \) (thus \( 2f = -2 \) or \( f = -1 \))
- Constant term is \( k \) (thus \( c = k \))

### Step 3: Apply the condition for a circle
For the equation to represent a circle, the following conditions must hold:
1. \( a = b \) (which is true since both are 1)
2. \( h = 0 \) (since there should be no \( xy \) term)

Thus, we find:
\[ h = 0 \]

### Step 4: Substitute values into the radius formula
The radius \( r \) of the circle is given by the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Given that the radius is 2, we can set up the equation:
\[ 2 = \sqrt{(-3)^2 + (-1)^2 - k} \]

### Step 5: Solve for \( k \)
Squaring both sides gives:
\[ 4 = 9 + 1 - k \]
\[ 4 = 10 - k \]
Rearranging gives:
\[ k = 10 - 4 = 6 \]

### Conclusion
Thus, the values of \( h \) and \( k \) are:
\[ h = 0, \quad k = 6 \]