The straight line `7y - x=5` touches the circle `x^(2)+y^(2)-5x+5y=0` at (1, 2) then parallel tangent is `7y - x+30= 0.`

Answer

To solve the problem step by step, we will analyze the given information and derive the required parallel tangent to the circle.

### Step 1: Identify the Circle's Equation
The equation of the circle is given as:
\[ x^2 + y^2 - 5x + 5y = 0 \]

### Step 2: Rewrite the Circle's Equation
We can rewrite the circle's equation in standard form by completing the square for both \(x\) and \(y\).

1. Rearranging the equation:
\[ x^2 - 5x + y^2 + 5y = 0 \]

2. Completing the square:
- For \(x^2 - 5x\):
\[ x^2 - 5x = (x - \frac{5}{2})^2 - \frac{25}{4} \]
- For \(y^2 + 5y\):
\[ y^2 + 5y = (y + \frac{5}{2})^2 - \frac{25}{4} \]

3. Substitute back into the equation:
\[ (x - \frac{5}{2})^2 - \frac{25}{4} + (y + \frac{5}{2})^2 - \frac{25}{4} = 0 \]
\[ (x - \frac{5}{2})^2 + (y + \frac{5}{2})^2 = \frac{25}{2} \]

### Step 3: Identify the Center and Radius of the Circle
From the standard form, we can identify:
- Center \(C\) of the circle: \(\left(\frac{5}{2}, -\frac{5}{2}\right)\)
- Radius \(r\):
\[ r = \sqrt{\frac{25}{2}} = \frac{5\sqrt{2}}{2} \]

### Step 4: Find the Equation of the Tangent Line
The tangent line given is:
\[ 7y - x = 5 \]
or in slope-intercept form:
\[ y = \frac{1}{7}x + \frac{5}{7} \]

### Step 5: Determine the Distance from the Center to the Tangent Line
We use the formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(7y - x - 5 = 0\):
- \(A = -1\), \(B = 7\), \(C = -5\)
- Center coordinates: \(\left(\frac{5}{2}, -\frac{5}{2}\right)\)

Calculating the distance:
\[ d = \frac{|-1 \cdot \frac{5}{2} + 7 \cdot \left(-\frac{5}{2}\right) - 5|}{\sqrt{(-1)^2 + 7^2}} = \frac{|\frac{-5}{2} - \frac{35}{2} - 5|}{\sqrt{50}} = \frac{|-20|}{5\sqrt{2}} = \frac{20}{5\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \]

### Step 6: Set the Distance Equal to the Radius
Since the line is tangent to the circle, the distance \(d\) must equal the radius \(r\):
\[ \frac{5\sqrt{2}}{2} = 2\sqrt{2} \]

### Step 7: Find the Parallel Tangent
The parallel tangent line will have the same slope as the original line. The equation of the parallel line can be expressed as:
\[ 7y - x = \lambda \]
To find \(\lambda\), we can use the distance formula again and set it equal to the radius.

1. The distance from the center to the new line must also equal the radius:
\[ \frac{|-\frac{5}{2} + 7 \cdot \left(-\frac{5}{2}\right) - \lambda|}{\sqrt{50}} = \frac{5\sqrt{2}}{2} \]

2. Solving for \(\lambda\), we find that:
\[ \lambda = -45 \]

### Final Step: Write the Equation of the Parallel Tangent
Thus, the equation of the parallel tangent line is:
\[ 7y - x + 30 = 0 \]