If `ax + by + c = 0` is the polar of `(1,1)` with respect to the circle `x^(2) + y^(2) - 2x + 2y `
`+ 1 = 0` and H. C. F. of a, b, c is equal to
one then find `a^(2) + b^(2) + c^(2) .`

Answer

To solve the problem, we need to find the values of \( a^2 + b^2 + c^2 \) given that the line \( ax + by + c = 0 \) is the polar of the point \( (1, 1) \) with respect to the circle defined by the equation \( x^2 + y^2 - 2x + 2y + 1 = 0 \).

### Step-by-Step Solution:

1. **Identify the Circle's Center and Radius:**
The given circle equation is:
\[ x^2 + y^2 - 2x + 2y + 1 = 0 \]
We can rewrite this in standard form. Completing the square for \( x \) and \( y \):
\[ (x - 1)^2 + (y + 1)^2 = 1^2 \]
Thus, the center of the circle is \( (1, -1) \) and the radius is \( 1 \).

2. **Find the Polar Line:**
The polar of a point \( (x_1, y_1) \) with respect to a circle is given by the equation:
\[ S_1 = 0 \]
where \( S = x^2 + y^2 - 2x + 2y + 1 \).

For the point \( (1, 1) \):
\[ S_1 = 1 \cdot 1 + 1 \cdot 1 - 2 \cdot 1 + 2 \cdot 1 + 1 = 1 + 1 - 2 + 2 + 1 = 3 \]
The polar line is given by:
\[ S_1 = 0 \implies x \cdot 1 + y \cdot 1 - (2 \cdot 1) + (2 \cdot 1) + 1 = 0 \]
This simplifies to:
\[ x + y - 2 + 2 + 1 = 0 \implies x + y + 1 = 0 \]

3. **Comparing Coefficients:**
The polar line we derived is:
\[ x + y + 1 = 0 \]
This can be compared with the general form \( ax + by + c = 0 \):
- Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \).

4. **Finding \( a^2 + b^2 + c^2 \):**
Now we calculate:
\[ a^2 + b^2 + c^2 = 1^2 + 1^2 + 1^2 = 1 + 1 + 1 = 3 \]

5. **Check the HCF Condition:**
The problem states that the HCF of \( a, b, c \) should be 1. Since \( \text{HCF}(1, 1, 1) = 1 \), this condition is satisfied.

### Final Answer:
Thus, the value of \( a^2 + b^2 + c^2 \) is:
\[ \boxed{3} \]