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The problem asks to find the equation of a line that is tangent to the circle given by the equation \( x^2 + y^2 - 2x + 8y + 4 = 0 \) and is parallel to the line \( x - y = 2 \).

### Step 1: Rewrite the equation of the circle in standard form.
We start with the given equation of the circle:

\[
x^2 + y^2 - 2x + 8y + 4 = 0
\]

Complete the square for both \(x\) and \(y\):

1. For \(x^2 - 2x\), we add and subtract \(\left(\frac{-2}{2}\right)^2 = 1\).
2. For \(y^2 + 8y\), we add and subtract \(\left(\frac{8}{2}\right)^2 = 16\).

The equation becomes:

\[
(x - 1)^2 + (y + 4)^2 = 13
\]

This represents a circle with center at \((1, -4)\) and radius \(\sqrt{13}\).

### Step 2: Find the slope of the tangent line.
The line \(x - y = 2\) has a slope of \(m = 1\), because it can be rewritten as \(y = x - 2\). Since the required tangent line is parallel to this line, it must also have a slope of 1.

### Step 3: Equation of the tangent line.
The equation of a line with slope \(1\) is:

\[
y = x + c
\]

where \(c\) is the y-intercept. Since this line is tangent to the circle, the distance from the center of the circle \((1, -4)\) to the line must be equal to the radius \(\sqrt{13}\).

### Step 4: Calculate the distance from the center to the line.
The formula for the distance from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is:

\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]

For the line \(x - y + c = 0\), we have \(A = 1\), \(B = -1\), and \(C = c\). The center of the circle is \((1, -4)\), so the distance is:

\[
d = \frac{|1(1) - 1(-4) + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|1 + 4 + c|}{\sqrt{2}} = \frac{|5 + c|}{\sqrt{2}}
\]

Set this equal to the radius \(\sqrt{13}\):

\[
\frac{|5 + c|}{\sqrt{2}} = \sqrt{13}
\]

Squaring both sides:

\[
\frac{(5 + c)^2}{2} = 13
\]

Multiply by 2:

\[
(5 + c)^2 = 26
\]

Take the square root of both sides:

\[
5 + c = \pm \sqrt{26}
\]

Thus:

\[
c = -5 \pm \sqrt{26}
\]

### Step 5: Final equations.
The two possible equations for the tangent lines are:

\[
y = x - 5 + \sqrt{26} \quad \text{or} \quad y = x - 5 - \sqrt{26}
\]

These are the equations of the lines tangent to the given circle and parallel to the line \(x - y = 2\).

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Would you like further details or have any questions?

Here are five related questions:
1. How do you find the point of tangency for each line?
2. What if the line wasn't parallel but perpendicular to the given line?
3. How would the procedure change for a different shaped curve like a parabola?
4. How can we calculate the distance between two circles?
5. Can we determine the area of the region between the tangent lines and the circle?

**Tip:** Completing the square is an essential tool for converting general conic equations into their standard forms, which is especially useful for identifying the center and radius of circles.

Tentukan persamaan garis singgung lingkaran x^2 + y^2 - 2x + 8y + 4 = 0 yang sejajar dengan garis x - y = 2.

This problem involves finding the equation of a tangent line to a circle given by x^2 + y^2 - 2x + 8y + 4 = 0, where the line is parallel to x - y = 2. The solution covers concepts in analytic geometry, including the distance from a point to a line and properties of tangent lines.

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