The equation of the tangent to the curve `x^(2)-2xy+y^(2)+2x+y-6=0` at (2,2) is

Options

A. `2x+y-6=0`
B. `2y+x-6=0`
C. `x+3y-8=0`
D. `3x+y-8=0`

Explanation

To find the equation of the tangent to the curve \( x^2 - 2xy + y^2 + 2x + y - 6 = 0 \) at the point \( (2, 2) \), we will follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation of the curve:
\[ x^2 - 2xy + y^2 + 2x + y - 6 = 0 \]
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(y) - \frac{d}{dx}(6) = 0 \]
This gives:
\[ 2x - 2\left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} + 2 + \frac{dy}{dx} = 0 \]

### Step 2: Collect terms involving \(\frac{dy}{dx}\)
Rearranging the equation, we have:
\[ 2x - 2y + 2 + \left(-2x + 2y + 1\right)\frac{dy}{dx} = 0 \]
Now, isolate \(\frac{dy}{dx}\):
\[ \left(-2x + 2y + 1\right)\frac{dy}{dx} = - (2x - 2y + 2) \]
Thus,
\[ \frac{dy}{dx} = \frac{-(2x - 2y + 2)}{-2x + 2y + 1} \]

### Step 3: Substitute the point (2, 2) to find the slope
Now we substitute \( x = 2 \) and \( y = 2 \):
\[ \frac{dy}{dx} = \frac{-(2(2) - 2(2) + 2)}{-2(2) + 2(2) + 1} \]
Calculating the numerator:
\[ -(4 - 4 + 2) = -2 \]
Calculating the denominator:
\[ -4 + 4 + 1 = 1 \]
Thus,
\[ \frac{dy}{dx} = \frac{-2}{1} = -2 \]

### Step 4: Write the equation of the tangent line
The slope of the tangent line at the point \( (2, 2) \) is \( -2 \). We use the point-slope form of the line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = -2 \), \( x_1 = 2 \), and \( y_1 = 2 \):
\[ y - 2 = -2(x - 2) \]
Expanding this:
\[ y - 2 = -2x + 4 \]
Rearranging gives:
\[ y = -2x + 6 \]

### Step 5: Final equation of the tangent line
We can also write this in standard form:
\[ 2x + y = 6 \]

### Conclusion
The equation of the tangent to the curve at the point \( (2, 2) \) is:
\[ \boxed{2x + y = 6} \]