What does the equation `(x - a)^(2) + ( y - b) ^(2) = c^(2)`
become when it is transferred to parallel axes through
the point (a, b - c) ? `

Answer

To solve the problem, we need to transform the given equation \((x - a)^{2} + (y - b)^{2} = c^{2}\) to a new coordinate system that is parallel to the original axes and shifted through the point \((a, b - c)\).

### Step-by-Step Solution:

1. **Identify the Original Equation:**
The original equation is given as:
\[ (x - a)^{2} + (y - b)^{2} = c^{2} \]

2. **Define the New Coordinate System:**
We need to shift the axes to a new point \((a, b - c)\). In the new coordinate system, we will denote the coordinates as \((x', y')\).

3. **Relate Old and New Coordinates:**
The relationship between the old coordinates \((x, y)\) and the new coordinates \((x', y')\) can be expressed as:
\[ x = x' + a \]
\[ y = y' + (b - c) \]

4. **Substitute the New Coordinates into the Original Equation:**
Now we substitute \(x\) and \(y\) in the original equation:
\[ (x' + a - a)^{2} + (y' + (b - c) - b)^{2} = c^{2} \]
This simplifies to:
\[ (x')^{2} + (y' - c)^{2} = c^{2} \]

5. **Expand the Equation:**
Expanding the left side gives:
\[ (x')^{2} + (y'^{2} - 2cy' + c^{2}) = c^{2} \]

6. **Simplify the Equation:**
Now, we can simplify the equation:
\[ (x')^{2} + y'^{2} - 2cy' + c^{2} = c^{2} \]
The \(c^{2}\) terms cancel out:
\[ (x')^{2} + y'^{2} - 2cy' = 0 \]

7. **Rearranging the Equation:**
Finally, we can rearrange the equation to get:
\[ (x')^{2} + y'^{2} = 2cy' \]

### Final Result:
The transformed equation in the new coordinate system is:
\[ (x')^{2} + y'^{2} = 2cy' \]