Find the slope of tangent to the curve if equation of the curve is `x^2+xy+y^2-2x-2y+1=0` at (0,0)

Answer

To find the slope of the tangent to the curve given by the equation \( x^2 + xy + y^2 - 2x - 2y + 1 = 0 \) at the point (0,0), we will follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation of the curve:
\[ x^2 + xy + y^2 - 2x - 2y + 1 = 0 \]
We will differentiate both sides with respect to \( x \).

### Step 2: Apply the differentiation rules
Using the rules of differentiation:
- The derivative of \( x^2 \) is \( 2x \).
- For \( xy \), we use the product rule: \( \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \).
- The derivative of \( y^2 \) is \( 2y \frac{dy}{dx} \).
- The derivative of \( -2x \) is \( -2 \).
- The derivative of \( -2y \) is \( -2 \frac{dy}{dx} \).
- The derivative of a constant (1) is 0.

Differentiating the entire equation gives:
\[ 2x + \left(x \frac{dy}{dx} + y\right) + \left(2y \frac{dy}{dx}\right) - 2 - 2 \frac{dy}{dx} = 0 \]

### Step 3: Rearranging the equation
Now, we can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ 2x + y + (x + 2y - 2) \frac{dy}{dx} - 2 = 0 \]
This simplifies to:
\[ (x + 2y - 2) \frac{dy}{dx} = 2 - 2x - y \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Now we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{2 - 2x - y}{x + 2y - 2} \]

### Step 5: Substitute the point (0,0)
We need to find the slope at the point (0,0):
\[ \frac{dy}{dx} \bigg|_{(0,0)} = \frac{2 - 2(0) - 0}{0 + 2(0) - 2} = \frac{2}{-2} = -1 \]

### Conclusion
The slope of the tangent to the curve at the point (0,0) is:
\[ \boxed{-1} \]