Find whether the straight line ` x + y = 2 + sqrt"" 2` touches the circle `x^(2) + y^(2) - 2x - 2y + 1 = 0`

Answer

To determine whether the line \(x + y = 2 + \sqrt{2}\) touches the circle given by the equation \(x^2 + y^2 - 2x - 2y + 1 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we will rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]
We can complete the square for both \(x\) and \(y\).

1. Rearranging the terms:
\[ (x^2 - 2x) + (y^2 - 2y) + 1 = 0 \]

2. Completing the square:
\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 + 1 = 0 \]
\[ (x - 1)^2 + (y - 1)^2 - 1 = 0 \]
\[ (x - 1)^2 + (y - 1)^2 = 1 \]

This shows that the circle has a center at \((1, 1)\) and a radius of \(1\).

### Step 2: Rewrite the Line's Equation
Next, we rewrite the line's equation:
\[ x + y = 2 + \sqrt{2} \]
We can express this in the standard form \(Ax + By + C = 0\):
\[ x + y - (2 + \sqrt{2}) = 0 \]

### Step 3: Calculate the Distance from the Center to the Line
We will use the formula for the distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \((x_0, y_0) = (1, 1)\), \(A = 1\), \(B = 1\), and \(C = -(2 + \sqrt{2})\).

Calculating the distance:
1. Substitute the values:
\[ d = \frac{|1 \cdot 1 + 1 \cdot 1 - (2 + \sqrt{2})|}{\sqrt{1^2 + 1^2}} \]
\[ = \frac{|1 + 1 - 2 - \sqrt{2}|}{\sqrt{2}} \]
\[ = \frac{|0 - \sqrt{2}|}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \]

### Step 4: Compare the Distance with the Radius
The radius of the circle is \(1\). Since the distance from the center of the circle to the line is equal to the radius, we conclude that the line touches the circle.

### Conclusion
The line \(x + y = 2 + \sqrt{2}\) touches the circle \(x^2 + y^2 - 2x - 2y + 1 = 0\).

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